Score : 500 points

Problem Statement

Alice and Brown loves games. Today, they will play the following game.

In this game, there are two piles initially consisting of X and Y stones, respectively.
Alice and Bob alternately perform the following operation, starting from Alice:

Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile.

The player who becomes unable to perform the operation, loses the game.

Given X and Y, determine the winner of the game, assuming that both players play optimally.

Constraints

0 ≤ X, Y ≤ 10^{18}

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the winner: either Alice or Brown.

Sample Input 1

2 1

Sample Output 1

Brown

Alice can do nothing but taking two stones from the pile containing two stones. As a result, the piles consist of zero and two stones, respectively. Then, Brown will take the two stones, and the piles will consist of one and zero stones, respectively. Alice will be unable to perform the operation anymore, which means Brown's victory.

Sample Input 2

5 0

Sample Output 2

Alice

Sample Input 3

0 0

Sample Output 3

Brown

Sample Input 4

4 8

Sample Output 4

Alice


#-swank
(unless (member :child-sbcl *features*)
  (quit
   :unix-status
   (process-exit-code
    (run-program *runtime-pathname*
                 `("--control-stack-size" "64MB"
                   "--noinform" "--disable-ldb" "--lose-on-corruption" "--end-runtime-options"
                   "--eval" "(push :child-sbcl *features*)"
                   "--script" ,(namestring *load-pathname*))
                 :output t :error t :input t))))
;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (ql:quickload '(:cl-debug-print :fiveam))
  #-swank (set-dispatch-macro-character #\# #\> (lambda (s c p) (declare (ignore c p)) (read s nil (values) t))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(defmacro with-buffered-stdout (&body body)
  "Buffers all outputs to *STANDARD-OUTPUT* in BODY and flushes them to
*STANDARD-OUTPUT* after BODY has been done (without error). Note that only
BASE-CHAR is allowed."
  (let ((out (gensym)))
    `(let ((,out (make-string-output-stream :element-type 'base-char)))
       (let ((*standard-output* ,out))
         ,@body)
       (write-string (get-output-stream-string ,out)))))

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48) (* 10 (the (integer 0 #.(floor most-positive-fixnum 10)) result))))
              (return (if minus (- result) result))))))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;; Body

(defun main ()
  (let* ((n (read))
         (m (read))
         (graph (make-array n :element-type 'list :initial-element nil))
         (table (make-array n :element-type 'bit :initial-element 0)))
    (declare (uint32 n m))
    (dotimes (i m)
      (let ((a (- (read-fixnum) 1))
            (b (- (read-fixnum) 1)))
        (push a (aref graph b))
        (push b (aref graph a))))
    (labels ((recur (v)
               (declare (uint32 v))
               (setf (aref table v) 1)
               (dolist (neighbor (aref graph v) (list v))
                 (when (zerop (aref table neighbor))
                   (return (cons v (recur neighbor)))))))
      (let* ((path1 (recur 0))
             (path2 (recur 0))
             (res (append (nreverse (cdr path1)) path2)))
        (println (length res))
        (with-buffered-stdout
          (loop with init = t
                for v in res
                do (if init
                       (setq init nil)
                       (write-char #\ ))
                   (write (+ v 1))))))))

#-swank (main)

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 500 points

Problem Statement

You are given a connected undirected simple graph, which has N vertices and M edges.
The vertices are numbered 1 through N, and the edges are numbered 1 through M.
Edge i connects vertices A_i and B_i.
Your task is to find a path that satisfies the following conditions:

The path traverses two or more vertices.

The path does not traverse the same vertex more than once.

A vertex directly connected to at least one of the endpoints of the path, is always contained in the path.

It can be proved that such a path always exists.
Also, if there are more than one solution, any of them will be accepted.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i < B_i \leq N

The given graph is connected and simple (that is, for every pair of vertices, there is at most one edge that directly connects them).

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Find one path that satisfies the conditions, and print it in the following format.
In the first line, print the count of the vertices contained in the path.
In the second line, print a space-separated list of the indices of the vertices, in order of appearance in the path.

Sample Input 1

5 6
1 3
1 4
2 3
1 5
3 5
2 4

Sample Output 1

4
2 3 1 4

There are two vertices directly connected to vertex 2: vertices 3 and 4.
There are also two vertices directly connected to vertex 4: vertices 1 and 2.
Hence, the path 2 → 3 → 1 → 4 satisfies the conditions.

Sample Input 2

7 8
1 2
2 3
3 4
4 5
5 6
6 7
3 5
2 6

Sample Output 2

7
1 2 3 4 5 6 7

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (progn (ql:quickload '(:cl-debug-print :fiveam))
                 (shadow :run)
                 (use-package :fiveam)))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

;; BEGIN_INSERTED_CONTENTS
(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  ;; (return-from read-fixnum 0)
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48) (* 10 (the (integer 0 #.(floor most-positive-fixnum 10)) result))))
              (return (if minus (- result) result))))))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;; Body

(declaim (inline count-collision))
(defun count-collision (anticlock clock l time)
  (if (< (* 2 time) (mod (- anticlock clock) l))
      0
      (let ((remain (- (* 2 time) (mod (- anticlock clock) l))))
        (+ 1 (floor remain l)))))

(defun main ()
  (declare #.OPT (inline sort))
  (let* ((n (read))
         (l (read))
         (time (read))
         (inits (make-array n :element-type 'uint32))
         (dirs (make-array n :element-type 'bit)))
    (declare (uint32 n l time))
    (dotimes (i n) (setf (aref inits i) (read-fixnum)
                         (aref dirs i) (mod (read-fixnum) 2)))
    (let* ((count0 (loop for i below n
                         unless (= (aref dirs 0) (aref dirs i))
                         sum (if (zerop (aref dirs 0))
                                 (count-collision (aref inits 0) (aref inits i) l time)
                                 (count-collision (aref inits i) (aref inits 0) l time))
                         of-type uint32))
           (end-index (mod (if (zerop (aref dirs 0))
                               (- count0)
                               count0)
                           n))
           (ends (make-array n :element-type 'uint32))
           (end0 (mod (+ (aref inits 0)
                         (if (zerop (aref dirs 0))
                             (- time)
                             time))
                      l)))
      (dotimes (i n)
        (setf (aref ends i)
              (mod (+ (aref inits i)
                      (if (zerop (aref dirs i))
                          (- time)
                          time))
                   l)))
      (setf ends (sort ends #'<))
      (let ((end0-pos (position end0 ends))
            (res (make-array n :element-type 'uint32)))
        (cond ((and (= (aref ends end0-pos) (aref ends (mod (+ end0-pos 1) n)))
                    (= 1 (aref dirs 0)))
               (setf end0-pos (mod (+ end0-pos 1) n)))
              ((and (= (aref ends end0-pos) (aref ends (mod (- end0-pos 1) n)))
                    (= 0 (aref dirs 0)))
               (setf end0-pos (mod (- end0-pos 1) n))))
        (let ((actual-index (mod (- end0-pos end-index) n)))
          (loop for i below n
                do (setf (aref res i)
                         (aref ends (mod (+ actual-index i) n))))
          (map () #'println res))))))

#-swank(main)

Score : 700 points

Problem Statement

There is a circle with a circumference of L.
Each point on the circumference has a coordinate value, which represents the arc length from a certain reference point clockwise to the point.
On this circumference, there are N ants.
These ants are numbered 1 through N in order of increasing coordinate, and ant i is at coordinate X_i.

The N ants have just started walking.
For each ant i, you are given the initial direction W_i. Ant i is initially walking clockwise if W_i is 1; counterclockwise if W_i is 2.
Every ant walks at a constant speed of 1 per second.
Sometimes, two ants bump into each other.
Each of these two ants will then turn around and start walking in the opposite direction.

For each ant, find its position after T seconds.

Constraints

All input values are integers.

1 \leq N \leq 10^5

1 \leq L \leq 10^9

1 \leq T \leq 10^9

0 \leq X_1 < X_2 < ... < X_N \leq L - 1

1 \leq W_i \leq 2

Input

The input is given from Standard Input in the following format:

N L T
X_1 W_1
X_2 W_2
:
X_N W_N

Output

Print N lines.
The i-th line should contain the coordinate of ant i after T seconds. Here, each coordinate must be between 0 and L-1, inclusive.

Sample Input 1

3 8 3
0 1
3 2
6 1

Sample Output 1

1
3
0

1.5 seconds after the ants start walking, ant 1 and 2 bump into each other at coordinate 1.5.
1 second after that, ant 1 and 3 bump into each other at coordinate 0.5.
0.5 seconds after that, that is, 3 seconds after the ants start walking, ants 1, 2 and 3 are at coordinates 1, 3 and 0, respectively.

Sample Input 2

4 20 9
7 2
9 1
12 1
18 1

Sample Output 2

7
18
18
1


#-swank
(unless (member :child-sbcl *features*)
  (quit
   :unix-status
   (process-exit-code
    (run-program *runtime-pathname*
                 `("--control-stack-size" "128MB"
                   "--noinform" "--disable-ldb" "--lose-on-corruption" "--end-runtime-options"
                   "--eval" "(push :child-sbcl *features*)"
                   "--script" ,(namestring *load-pathname*))
                 :output t :error t :input t))))
(in-package :cl-user)
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx opt
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(defconstant +mod+ 1000000007)

(defmacro dbg (&rest forms)
  #+swank (if (= (length forms) 1)
              `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
              `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

;; BEGIN_INSERTED_CONTENTS
(defpackage :cp/println-matrix
  (:use :cl)
  (:export #:println-matrix))
(in-package :cp/println-matrix)

(declaim (inline println-matrix))
(defun println-matrix (array &key (separator #\ ) (key #'identity) (writer #'write) (row-start 0) row-end (col-start 0) col-end)
  "Prints a 2-dimensional array."
  (declare ((array * (* *)) array)
           ((integer 0 #.most-positive-fixnum) row-start col-start))
  (let ((row-end (or row-end (array-dimension array 0)))
        (col-end (or col-end (array-dimension array 1))))
    (declare ((integer 0 #.most-positive-fixnum) row-end col-end))
    (loop for i from row-start below row-end
          do (loop for j from col-start below col-end
                   unless (= j col-start)
                   do (princ separator)
                   do (funcall writer (funcall key (aref array i j))))
             (terpri))))

;; BEGIN_USE_PACKAGE
(eval-when (:compile-toplevel :load-toplevel :execute)
  (use-package :cp/println-matrix :cl-user))
(in-package :cl-user)

;;;
;;; Body
;;;

(defun test (n m &rest as)
  (let ((matrix (make-array (list n 7) :element-type 'base-char :initial-element #\.)))
    (dotimes (i (* n 7))
      (when (member (mod i m) as)
        (setf (row-major-aref matrix i) #\#)))
    (println-matrix matrix :separator "" :writer #'write-char)))

(defun solve-small (n m as)
  (declare (uint31 n m)
           ((simple-array uint31 (*)) as))
  (let ((plan (make-array (list n 7) :element-type 'bit :initial-element 0))
        (res 0))
    (declare (uint62 res))
    (loop for a across as
          do (loop for i from a below (* n 7) by m
                   do (setf (row-major-aref plan i) 1)))
    (dotimes (i n)
      (dotimes (j 7)
        (when (zerop (aref plan i j))
          (sb-int:named-let dfs ((i i) (j j))
            (labels ((visit (y x)
                       (when (and (<= 0 y (- n 1))
                                  (<= 0 x 6))
                         (dfs y x))))
              (when (zerop (aref plan i j))
                (setf (aref plan i j) 1)
                (visit (- i 1) j)
                (visit (+ i 1) j)
                (visit i (- j 1))
                (visit i (+ j 1)))))
          (incf res))))
    res))

(defun main ()
  (let* ((n (read))
         (m (read))
         (q (read))
         (as (make-array q :element-type 'uint31 :initial-element 0)))
    (dotimes (i q)
      (setf (aref as i) (read-fixnum)))
    (println
     (cond
       ((zerop (mod m 7))
        (let* ((period (floor m 7))
               (term1 (solve-small period m as))
               (term2 (solve-small (* 2 period) m as))
               (coef (- term2 term1))
               (intercept (- term1 coef)))
          (multiple-value-bind (quot rem) (floor n period)
            (dbg coef intercept)
            (+ intercept (* quot coef)
               (- (solve-small (+ period rem) m as) term1)))))
       ((<= n 10000)
        (solve1 n m as))
       (t (error "Huh?"))))))

#-swank (main)

;;;
;;; Test and benchmark
;;;

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    #+os-windows (run-program "powershell.exe" '("-Command" "Get-Clipboard") :output out :search t)
    #+os-unix (run-program "xsel" '("-b" "-o") :output out :search t)))

#+swank (defparameter *this-pathname* (uiop:current-lisp-file-pathname))
#+swank (defparameter *dat-pathname* (uiop:merge-pathnames* "test.dat" *this-pathname*))

#+swank
(defun run (&optional thing (out *standard-output*))
  "THING := null | string | symbol | pathname

null: run #'MAIN using the text on clipboard as input.
string: run #'MAIN using the string as input.
symbol: alias of FIVEAM:RUN!.
pathname: run #'MAIN using the text file as input."
  (let* ((*standard-output* (or out (make-string-output-stream)))
         (res (etypecase thing
                (null
                 (with-input-from-string (*standard-input* (delete #\Return (get-clipbrd)))
                   (main)))
                (string
                 (with-input-from-string (*standard-input* (delete #\Return thing))
                   (main)))
                (symbol (5am:run! thing))
                (pathname
                 (with-open-file (*standard-input* thing)
                   (main))))))
    (if out res (get-output-stream-string *standard-output*))))

#+swank
(defun gen-dat ()
  (uiop:with-output-file (out *dat-pathname* :if-exists :supersede)
    (format out "")))

#+swank
(defun bench (&optional (out (make-broadcast-stream)))
  (time (run *dat-pathname* out)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (equal "4
"
          (run "7 7 3
1 3 5
" nil)))
  (it.bese.fiveam:is
   (equal "10
"
          (run "10 14 8
5 6 7 8 9 10 11 12
" nil))))

Max Score: 500 Points

Problem Statement

We have a grid with n rows and 7 columns. We call it a calendar. The cell at i-th row and j-th column is denoted (i, j).

Initially, each cell at (i, j) contains the integer 7i + j - 8, and each cell is white.

Snuke likes painting, so he decided integer m, and did q operations with a calendar.

・In i-th operation, he paint black on the cell in which an integer is written such remainder of dividing by m is a_i.

Please count the number of connected white parts.

Note that if two adjacent cells are white, the cells belong to the same connected part.

Input Format

The input format is following:

n m q
a_1 a_2 ... a_q

Output Format

Print the number of connected part in one line.

Constraints

n ≤ 10^{12}

7n is divisible by m.

1 ≤ q ≤ m ≤ 10^5

0 ≤ a_1 < a_2 < ... < a_q < m

Scoring

Subtask 1 [100 points]

n ≤ 100000.

Subtask 2 [90 points]

m is divisible by 7.

a_{i + 1} - a_i = 1.

Subtask 3 [200 points]

m is divisible by 7.

Subtask 4 [110 points]

There are no additional constraints.

Sample Input 1

7 7 3
1 3 5

Sample Output 1

4

The calendar looks like this:

Sample Input 2

10 14 8
5 6 7 8 9 10 11 12

Sample Output 2

10

The calendar looks like this:


(let ((a (read))
      (b (read))
      (c (read)))

  (format t "~A~%"
          (if (= (- b a) (- c b))
            'yes
            'no)))

Score : 100 points

Problem Statement

Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right.
We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b-a = c-b.

Determine whether the arrangement of the poles is beautiful.

Constraints

1 \leq a,b,c \leq 100

a, b and c are integers.

Input

Input is given from Standard Input in the following format:

a b c

Output

Print YES if the arrangement of the poles is beautiful; print NO otherwise.

Sample Input 1

2 4 6

Sample Output 1

YES

Since 4-2 = 6-4, this arrangement of poles is beautiful.

Sample Input 2

2 5 6

Sample Output 2

NO

Since 5-2 \neq 6-5, this arrangement of poles is not beautiful.

Sample Input 3

3 2 1

Sample Output 3

YES

Since 1-2 = 2-3, this arrangement of poles is beautiful.


(defun solver ()
  (let ((a (read)) (b (read)) (c (read)))
  (if (= (- b a) (- c b))
      (format t "YES~%")
      (format t "NO~%"))))

(solver)


(format t "~{~A~}~%" (loop for odds across (read-line)
                       for evens across (concatenate 'string (read-line) " ")
                       append (list odds evens)))

Score : 200 points

Problem Statement

Snuke signed up for a new website which holds programming competitions.
He worried that he might forget his password, and he took notes of it.
Since directly recording his password would cause him trouble if stolen,
he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.

You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order.
Restore the original password.

Constraints

O and E consists of lowercase English letters (a - z).

1 \leq |O|,|E| \leq 50

|O| - |E| is either 0 or 1.

Input

Input is given from Standard Input in the following format:

O
E

Output

Print the original password.

Sample Input 1

xyz
abc

Sample Output 1

xaybzc

The original password is xaybzc. Extracting the characters at the odd-numbered positions results in xyz, and extracting the characters at the even-numbered positions results in abc.

Sample Input 2

atcoderbeginnercontest
atcoderregularcontest

Sample Output 2

aattccooddeerrbreeggiunlnaerrccoonntteesstt


(let* ((si (loop repeat (read)
                 collect (coerce (read-line) 'list)))
       (a (make-array 26 :initial-element 50)))
 (mapc (lambda (s)
         (let ((b (make-array 26 :initial-element 0)))
           (mapc (lambda (c)
                   (incf (aref b (- (char-code c) 97))))
               s)
           (loop for y across b
                 for i from 0
                 when (> (aref a i) y)
                 do (setf (aref a i) y))))
       si)
 (loop for x across a
       for i from 97
       do (loop repeat x
                do (princ (code-char i))))
 (fresh-line))

Score : 300 points

Problem Statement

Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string.

He will receive a headline which contains one of the strings S_1,...,S_n tomorrow.
He is excited and already thinking of what string he will create.
Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains.

Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains.
If there are multiple such strings, find the lexicographically smallest one among them.

Constraints

1 \leq n \leq 50

1 \leq |S_i| \leq 50 for every i = 1, ..., n.

S_i consists of lowercase English letters (a - z) for every i = 1, ..., n.

Input

Input is given from Standard Input in the following format:

n
S_1
...
S_n

Output

Print the lexicographically smallest string among the longest strings that satisfy the condition.
If the answer is an empty string, print an empty line.

Sample Input 1

3
cbaa
daacc
acacac

Sample Output 1

aac

The strings that can be created from each of cbaa, daacc and acacac, are aa, aac, aca, caa and so forth.
Among them, aac, aca and caa are the longest, and the lexicographically smallest of these three is aac.

Sample Input 2

3
a
aa
b

Sample Output 2

The answer is an empty string.


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (progn (ql:quickload '(:cl-debug-print :fiveam))
                 (shadow :run)
                 (use-package :fiveam)))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

;; BEGIN_INSERTED_CONTENTS
(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  ;; (return-from read-fixnum 0)
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48) (* 10 (the (integer 0 #.(floor most-positive-fixnum 10)) result))))
              (return (if minus (- result) result))))))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;; Body

(defun main ()
  (let* ((n (read))
         (m (read))
         (xs (make-array n :element-type 'int32))
         (ys (make-array m :element-type 'int32)))
    (declare ((integer 0 100000) n m))
    (dotimes (j n) (setf (aref xs j) (read-fixnum)))
    (dotimes (i m) (setf (aref ys i) (read-fixnum)))
    (println
     (mod
      (* (loop with res of-type uint32 = 0
               for i from 1 to (- m 1)
               do (setf res (mod (+ res (* i
                                           (- m i)
                                           (- (aref ys i) (aref ys (- i 1)))))
                                 +mod+))
               finally (return res))
         (loop with res of-type uint32 = 0
               for j from 1 to (- n 1)
               do (setf res (mod (+ res (* j
                                           (- n j)
                                           (- (aref xs j) (aref xs (- j 1)))))
                                 +mod+))
               finally (return res)))
      +mod+))))

#-swank(main)

Score : 500 points

Problem Statement

On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis.
Among the lines parallel to the x axis, the i-th from the bottom is represented by y = y_i.
Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x = x_i.

For every rectangle that is formed by these lines, find its area, and print the total area modulo 10^9+7.

That is, for every quadruple (i,j,k,l) satisfying 1\leq i < j\leq n and 1\leq k < l\leq m, find the area of the rectangle formed by the lines x=x_i, x=x_j, y=y_k and y=y_l, and print the sum of these areas modulo 10^9+7.

Constraints

2 \leq n,m \leq 10^5

-10^9 \leq x_1 < ... < x_n \leq 10^9

-10^9 \leq y_1 < ... < y_m \leq 10^9

x_i and y_i are integers.

Input

Input is given from Standard Input in the following format:

n m
x_1 x_2 ... x_n
y_1 y_2 ... y_m

Output

Print the total area of the rectangles, modulo 10^9+7.

Sample Input 1

3 3
1 3 4
1 3 6

Sample Output 1

60

The following figure illustrates this input:

The total area of the nine rectangles A, B, ..., I shown in the following figure, is 60.

Sample Input 2

6 5
-790013317 -192321079 95834122 418379342 586260100 802780784
-253230108 193944314 363756450 712662868 735867677

Sample Output 2

835067060


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) (read s nil (values) t))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48) (* 10 (the (integer 0 #.(floor most-positive-fixnum 10)) result))))
              (return (if minus (- result) result))))))))

;; Should we do this with UNWIND-PROTECT?
(defmacro with-buffered-stdout (&body body)
  "Buffers all outputs to *STANDARD-OUTPUT* in BODY and flushes them to
*STANDARD-OUTPUT* after BODY has been done (without error). Note that only
BASE-CHAR is allowed."
  (let ((out (gensym)))
    `(let ((,out (make-string-output-stream :element-type 'base-char)))
       (let ((*standard-output* ,out))
         ,@body)
       (write-string (get-output-stream-string ,out)))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun main ()
  (declare #.OPT)
  (let* ((ss (read-line))
         (tt (read-line))
         (slen (length ss))
         (tlen (length tt))
         (cumuls (make-array (+ 1 slen) :element-type 'uint32 :initial-element 0))
         (cumult (make-array (+ 1 tlen) :element-type 'uint32 :initial-element 0))
         (q (read)))
    (declare (simple-string ss tt)
             (uint32 q))
    (dotimes (i slen)
      (setf (aref cumuls (+ i 1))
            (+ (aref cumuls i) (if (char= #\A (aref ss i)) 1 2))))
    (dotimes (i tlen)
      (setf (aref cumult (+ i 1))
            (+ (aref cumult i) (if (char= #\A (aref tt i)) 1 2))))
    (with-buffered-stdout
      (dotimes (i q)
        (let ((a (- (read-fixnum) 1))
              (b (read-fixnum))
              (c (- (read-fixnum) 1))
              (d (read-fixnum)))
          (if (= (mod (- (aref cumuls b) (aref cumuls a)) 3)
                 (mod (- (aref cumult d) (aref cumult c)) 3))
              (write-line "YES")
              (write-line "NO")))))))

#-swank (main)

Score : 600 points

Problem Statement

Let us consider the following operations on a string consisting of A and B:

Select a character in a string. If it is A, replace it with BB. If it is B, replace with AA.

Select a substring that is equal to either AAA or BBB, and delete it from the string.

For example, if the first operation is performed on ABA and the first character is selected, the string becomes BBBA.
If the second operation is performed on BBBAAAA and the fourth through sixth characters are selected, the string becomes BBBA.

These operations can be performed any number of times, in any order.

You are given two string S and T, and q queries a_i, b_i, c_i, d_i.
For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.

Constraints

1 \leq |S|, |T| \leq 10^5

S and T consist of letters A and B.

1 \leq q \leq 10^5

1 \leq a_i \leq b_i \leq |S|

1 \leq c_i \leq d_i \leq |T|

Input

Input is given from Standard Input in the following format:

S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q

Output

Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print YES. Otherwise, print NO.

Sample Input 1

BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4

Sample Output 1

YES
NO
YES
NO

The first query asks whether the string ABA can be made into BBBA.
As explained in the problem statement, it can be done by the first operation.

The second query asks whether ABA can be made into BBBB, and the fourth query asks whether BBBAAAA can be made into BBB.
Neither is possible.

The third query asks whether the string BBBAAAA can be made into BBBA.
As explained in the problem statement, it can be done by the second operation.

Sample Input 2

AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14

Sample Output 2

YES
YES
YES
YES
YES
YES
NO
NO
NO
NO


(defun input (n)
  (let ((A (make-array n)))
    (loop for i below n do (setf (aref A i) (read)))
    A))
(defun solve (n A)
  (loop for i from n below (* 2 n) sum (aref A i)))
(let* ((n (read))
       (A (input (* 3 n))))
  (sort A #'>)
  (princ (solve n A)))

Score : 300 points

Problem Statement

There are 3N participants in AtCoder Group Contest.
The strength of the i-th participant is represented by an integer a_i.
They will form N teams, each consisting of three participants.
No participant may belong to multiple teams.

The strength of a team is defined as the second largest strength among its members.
For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.

Find the maximum possible sum of the strengths of N teams.

Constraints

1 ≤ N ≤ 10^5

1 ≤ a_i ≤ 10^{9}

a_i are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the answer.

Sample Input 1

2
5 2 8 5 1 5

Sample Output 1

10

The following is one formation of teams that maximizes the sum of the strengths of teams:

Team 1: consists of the first, fourth and fifth participants.

Team 2: consists of the second, third and sixth participants.

Sample Input 2

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

10000000000

The sum of the strengths can be quite large.


(eval-when  (:compile-toplevel :load-toplevel :execute)
  (defun script-p ()
    (eql sb-sys:*stdin* *standard-input*)))

(macrolet ((optimize (pred)
	     (if pred
		 `(declaim (optimize (speed 3) (debug 0) (safety 0)))
		 `(declaim (optimize (speed 0) (debug 3) (safety 3))))))
  (optimize #. (script-p)))

(defun xclip ()
  (with-output-to-string (s)
    (sb-ext:run-program "/usr/bin/xclip" '("-o") :output s)))

(defun test ()
  (with-input-from-string (*standard-input* (xclip))
    (main)))

(defun main ()
  (let* ((n (read))
	 (as (sort (loop :repeat (* 3 n)
		      :collect (read))
		   #'>)))
    (format t "~a~%"
	    (loop :repeat n
	       :for x :in (cdr as) :by #'cddr
	       :sum x))))

(when (script-p) (main))


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (ql:quickload '(:cl-debug-print :fiveam))
  #-swank (set-dispatch-macro-character #\# #\> (lambda (s c p) (declare (ignore c p)) (read s nil (values) t))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(defmacro with-buffered-stdout (&body body)
  "Buffers all outputs to *STANDARD-OUTPUT* in BODY and flushes them to
*STANDARD-OUTPUT* after BODY has been done (without error). Note that only
BASE-CHAR is allowed."
  (let ((out (gensym)))
    `(let ((,out (make-string-output-stream :element-type 'base-char)))
       (let ((*standard-output* ,out))
         ,@body)
       (write-string (get-output-stream-string ,out)))))

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48) (* 10 (the (integer 0 #.(floor most-positive-fixnum 10)) result))))
              (return (if minus (- result) result))))))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;; Body

(defun main ()
  (let* ((n (read))
         (m (read))
         (graph (make-array n :element-type 'list :initial-element nil))
         (colors (make-array n :element-type 'int32 :initial-element -1))
         (dp (make-array n :element-type 'int32 :initial-element 0)))
    (declare (uint32 n m))
    (dotimes (i m)
      (let ((a (- (read-fixnum) 1))
            (b (- (read-fixnum) 1)))
        (push a (aref graph b))
        (push b (aref graph a))))
    (let ((q (read)))
      ;; reverse ordering
      (sb-int:named-let recur ((count 0))
        (unless (= q count)
          (let ((v (- (read-fixnum) 1))
                (d (read-fixnum))
                (c (read-fixnum)))
            (recur (+ count 1))
            (sb-int:named-let dfs ((v v) (depth d))
              ;; if not colored, color it
              (when (= -1 (aref colors v))
                (setf (aref colors v) c))
              ;; search neighbors if the current depth is larger than the
              ;; maximal old depth.
              (when (> depth (aref dp v))
                (setf (aref dp v) depth)
                (dolist (neighbor (aref graph v))
                  (dfs neighbor (- depth 1)))))))))
    (with-buffered-stdout
      (dotimes (i n)
        (println (max 0 (aref colors i)))))))

#-swank (main)

Score : 700 points

Problem Statement

Squid loves painting vertices in graphs.

There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges.
Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.

Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.

Find the color of each vertex after the Q operations.

Constraints

1 ≤ N,M,Q ≤ 10^5

1 ≤ a_i,b_i,v_i ≤ N

a_i ≠ b_i

0 ≤ d_i ≤ 10

1 ≤ c_i ≤10^5

d_i and c_i are all integers.

There are no self-loops or multiple edges in the given graph.

Partial Score

200 points will be awarded for passing the testset satisfying 1 ≤ N,M,Q ≤ 2{,}000.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_{M} b_{M}
Q
v_1 d_1 c_1
:
v_{Q} d_{Q} c_{Q}

Output

Print the answer in N lines.
In the i-th line, print the color of vertex i after the Q operations.

Sample Input 1

7 7
1 2
1 3
1 4
4 5
5 6
5 7
2 3
2
6 1 1
1 2 2

Sample Output 1

2
2
2
2
2
1
0

Initially, each vertex is painted in color 0.
In the first operation, vertices 5 and 6 are repainted in color 1.
In the second operation, vertices 1, 2, 3, 4 and 5 are repainted in color 2.

Sample Input 2

14 10
1 4
5 7
7 11
4 10
14 7
14 3
6 14
8 11
5 13
8 3
8
8 6 2
9 7 85
6 9 3
6 7 5
10 3 1
12 9 4
9 6 6
8 2 3

Sample Output 2

1
0
3
1
5
5
3
3
6
1
3
4
5
3

The given graph may not be connected.


(eval-when  (:compile-toplevel :load-toplevel :execute)
  (defun script-p ()
    (eql sb-sys:*stdin* *standard-input*)))

;;; helper
(defmacro defvars (&rest vars)
  `(progn ,@(mapcar #'(lambda (x) `(defvar ,x)) vars)))

(macrolet ((optimize (pred)
	     (if pred
		 `(declaim (optimize (speed 3) (debug 0) (safety 0)))
		 `(declaim (optimize (speed 0) (debug 3) (safety 3))))))
  (optimize #. (script-p)))

(defun xclip ()
  (with-output-to-string (s)
    (sb-ext:run-program "/usr/bin/xclip" '("-sel" "clip" "-o") :output s)))

(defun test ()
  (with-input-from-string (*standard-input* (xclip))
    (main)))

(defmacro show (name)
  (if (script-p)
      nil
      `(format t ,(concatenate 'string (format nil "~a" name) " is ~a~%")
	       ,name)))

;;; code starts here
(defparameter dmax 10)

(defstruct v color ns mark)

(defvars n m vs l dp init colors)

;; table[v][d]
;;  v: vertex
;;  d: distace
;;  value: -1 invalid
;;       |  0 not colored
;;       |  n otherwise

(defun rundp (v d)
  (cond ((> d dmax) 0)
	((plusp (aref dp v d)) (aref dp v d))
	(t (setf (aref dp v d)
		 (max (aref init v d)
		      (loop for x in (aref vs v)
			 maximize (rundp x (1+ d))))))))

(defun main ()
  (setf n (read))
  (setf m (read))
  (setf vs (make-array n :initial-element nil))

  (loop for i from 0 below n
       do (push i (aref vs i)))

  (loop repeat m
     for a = (1- (read))
     for b = (1- (read))
     do (push b (aref vs a))
     do (push a (aref vs b)))

  (show vs)

  (setf l (read))
  (setf dp (make-array (list n (1+ dmax)) :initial-element -1))
  (setf init (make-array (list n (1+ dmax)) :initial-element -1))
  (setf colors (make-array (1+ l)))

  (loop repeat l
     for i from 1 ;; 1!
     for v = (1- (read))
     for d = (read)
     for c = (read)
     do (setf (aref init v d) i
	      (aref colors i) c))

  (show init)
  (show colors)

  (loop for v from 0 below n
     do (format t "~a~%" (aref colors (rundp v 0)))))

(when (script-p) (main))


(format t "~A~%"
        (mod (+ (read) (read)) 24))

Score : 100 points

Problem Statement

Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.

In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".

The current time is A o'clock, and a contest will begin in exactly B hours.
When will the contest begin? Answer in 24-hour time.

Constraints

0 \leq A,B \leq 23

A and B are integers.

Input

The input is given from Standard Input in the following format:

A B

Output

Print the hour of the starting time of the contest in 24-hour time.

Sample Input 1

9 12

Sample Output 1

21

In this input, the current time is 9 o'clock, and 12 hours later it will be 21 o'clock in 24-hour time.

Sample Input 2

19 0

Sample Output 2

19

The contest has just started.

Sample Input 3

23 2

Sample Output 3

1

The contest will begin at 1 o'clock the next day.


(let ((a (read)) (b (read)))
      (princ (mod (+ a b) 24)))


(defun split-with (string &key (delimiterp #'(lambda (c) (char= c #\Space))))
  (loop :for beg = (position-if-not delimiterp string)
        :then (position-if-not delimiterp string :start (1+ end))
        :for end = (and beg (position-if delimiterp string :start beg))
        :when beg :collect (subseq string beg end)
        :while end))

(let* ((lst (split-with (read-line)))
       (a (parse-integer (car lst)))
       (b (parse-integer (car (cdr lst)))))
  (format t "~d~%" (mod (+ a b) 24)))


(let* ((n (read))
       (m (read))
       (stu (loop :for k :from 1 :upto n :collect (list (read) (read))))
       (ckpt (loop :for k :from 1 :upto m :collect (list k (read) (read)))))
  (defun dist (a b)
    (+ (abs (- (first a) (first b))) (abs (- (second a) (second b)))))
  (defun f (k)
    (let* ((ans 1))
      (loop :for x :from 1 :upto m :do(if (> (dist k (cdr (nth (1- ans) ckpt))) (dist k (cdr (nth (1- x) ckpt)))) (setf ans x)))
      ans))
  )

Score : 200 points

Problem Statement

There are N students and M checkpoints on the xy-plane.

The coordinates of the i-th student (1 \leq i \leq N) is (a_i,b_i), and the coordinates of the checkpoint numbered j (1 \leq j \leq M) is (c_j,d_j).

When the teacher gives a signal, each student has to go to the nearest checkpoint measured in Manhattan distance.

The Manhattan distance between two points (x_1,y_1) and (x_2,y_2) is |x_1-x_2|+|y_1-y_2|.

Here, |x| denotes the absolute value of x.

If there are multiple nearest checkpoints for a student, he/she will select the checkpoint with the smallest index.

Which checkpoint will each student go to?

Constraints

1 \leq N,M \leq 50

-10^8 \leq a_i,b_i,c_j,d_j \leq 10^8

All input values are integers.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_N b_N
c_1 d_1
:
c_M d_M

Output

Print N lines.

The i-th line (1 \leq i \leq N) should contain the index of the checkpoint for the i-th student to go.

Sample Input 1

2 2
2 0
0 0
-1 0
1 0

Sample Output 1

2
1

The Manhattan distance between the first student and each checkpoint is:

For checkpoint 1: |2-(-1)|+|0-0|=3

For checkpoint 2: |2-1|+|0-0|=1

The nearest checkpoint is checkpoint 2. Thus, the first line in the output should contain 2.

The Manhattan distance between the second student and each checkpoint is:

For checkpoint 1: |0-(-1)|+|0-0|=1

For checkpoint 2: |0-1|+|0-0|=1

When there are multiple nearest checkpoints, the student will go to the checkpoint with the smallest index. Thus, the second line in the output should contain 1.

Sample Input 2

3 4
10 10
-10 -10
3 3
1 2
2 3
3 5
3 5

Sample Output 2

3
1
2

There can be multiple checkpoints at the same coordinates.

Sample Input 3

5 5
-100000000 -100000000
-100000000 100000000
100000000 -100000000
100000000 100000000
0 0
0 0
100000000 100000000
100000000 -100000000
-100000000 100000000
-100000000 -100000000

Sample Output 3

5
4
3
2
1


(let ((n (read))
    (ans 0)
    (a 0)
    (b 0))
    (loop for i from 1 while (<= (* i i) n) do
        (if (zerop (rem n i))
        (progn
            (setq a i)
            (setq b (/ n i))
            (if (= ans 0)
                (setq ans (max (length (princ-to-string a)) (length (princ-to-string b))))
                (if (> ans (max (length (princ-to-string a)) (length (princ-to-string b))))
                    (setq ans (max (length (princ-to-string a)) (length (princ-to-string b))))
                )
            )
            (if (not (= (* i i) n))
                (progn
                    (setq a i)
                    (setq b (/ n i))
                    (if (> ans (max (length (princ-to-string a)) (length (princ-to-string b))))
                    (setq ans (max (length (princ-to-string a)) (length (princ-to-string b))))
                )
                )
            )
        )
        )
    )
    (princ ans)
)

Score : 300 points

Problem Statement

You are given an integer N.

For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.

For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.

Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

1 \leq N \leq 10^{10}

N is an integer.

Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (progn (ql:quickload '(:cl-debug-print :fiveam))
                 (shadow :run)
                 (use-package :fiveam))
  #-swank (set-dispatch-macro-character #\# #\> (lambda (s c p) (declare (ignore c p)) (read s nil (values) t))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

;; BEGIN_INSERTED_CONTENTS
;;;
;;; Compute binomial coefficient by direct bignum arithmetic
;;;
;;; This code ist almost the same as that of alexandria.
;;;

(declaim (inline %multiply-range))
(defun %multiply-range (i j)
  (labels ((bisect (j k)
             (declare (type (integer 1 #.most-positive-fixnum) j k)
                      (values integer))
             (if (< (- k j) 8)
                 (multiply-range j k)
                 (let ((middle (+ j (truncate (- k j) 2))))
                   (* (bisect j middle)
                      (bisect (+ middle 1) k)))))
           (multiply-range (j k)
             (declare (type (integer 1 #.most-positive-fixnum) j k))
             (do ((f k (* f m))
                  (m (1- k) (1- m)))
                 ((< m j) f)
               (declare (type (integer 0 (#.most-positive-fixnum)) m)
                        (type unsigned-byte f)))))
    (bisect i j)))

(declaim (inline factorial))
(defun factorial (n)
  (%multiply-range 1 n))

(defun binomial-coefficient (n k)
  (declare #.OPT
           ((integer 0 (#.most-positive-fixnum)) n k))
  (assert (>= n k))
  (if (or (zerop k) (= n k))
      1
      (let ((n-k (- n k)))
        (when (< k n-k)
          (rotatef k n-k))
        (if (= 1 n-k)
            n
            (floor (%multiply-range (+ k 1) n)
	           (%multiply-range 1 n-k))))))

(defun multiset-coefficient (n k)
  (binomial-coefficient (+ n k -1) k))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;; Body

(defun main ()
  (declare #.OPT)
  (let* ((n (read))
         (a (read))
         (b (read))
         (vs (sort (make-array n :element-type 'uint62 :initial-contents (the list (loop repeat n collect (read)))) #'>))
         (max-score (loop for end from a to b maximize (/ (loop for i below end sum (aref vs i) of-type uint62) end)))
         (prevs (make-array n :element-type 'uint32 :initial-element 1)))
    (declare (uint8 n a b))
    (println (float max-score 1d0))
    (loop for i from 1 below n
          do (if (= (aref vs i) (aref vs (- i 1)))
                 (setf (aref prevs i) (+ (aref prevs (- i 1)) 1))
                 1))
    (let ((chunks (copy-seq prevs))
          (i (- n 1)))
      (declare ((simple-array uint32 (*)) chunks))
      (loop (when (<= i 0)
              (return))
            (loop for j from i downto 0
                  until (= 1 (aref chunks j))
                  do (setf (aref chunks j) (aref chunks i))
                  finally (setf (aref chunks j) (aref chunks i))
                          (setf i (- j 1))))
      (println
       (loop for end from a to b
             for score = (/ (loop for i below end sum (aref vs i) of-type uint62) end)
             when (= score max-score)
             sum (binomial-coefficient (aref chunks (- end 1)) (aref prevs (- end 1)))
             of-type uint62)))))

#-swank(main)

Score : 400 points

Problem Statement

You are given N items.

The value of the i-th item (1 \leq i \leq N) is v_i.

Your have to select at least A and at most B of these items.

Under this condition, find the maximum possible arithmetic mean of the values of selected items.

Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

1 \leq N \leq 50

1 \leq A,B \leq N

1 \leq v_i \leq 10^{15}

Each v_i is an integer.

Input

The input is given from Standard Input in the following format:

N A B
v_1
v_2
...
v_N

Output

Print two lines.

The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.

The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Sample Input 1

5 2 2
1 2 3 4 5

Sample Output 1

4.500000
1

The mean of the values of selected items will be maximized when selecting the fourth and fifth items. Hence, the first line of the output should contain 4.5.

There is no other way to select items so that the mean of the values will be 4.5, and thus the second line of the output should contain 1.

Sample Input 2

4 2 3
10 20 10 10

Sample Output 2

15.000000
3

There can be multiple ways to select items so that the mean of the values will be maximized.

Sample Input 3

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996

Sample Output 3

1000000000000000.000000
1


(defun input (n)
  (let ((a (make-array n)))
    (loop for i below n do (setf (aref a i) (read)))
    a))

(defparameter *fact-array* (make-array 60))
(setf (aref *fact-array* 0) 1)
(setf (aref *fact-array* 1) 1)
(defun fact (n)
  (let ((x (aref *fact-array* n)))
    (if (> x 0)
      x
      (progn (setf (aref *fact-array* n) (* n (fact (1- n))))
             (aref *fact-array* n)))))
(defun ncr (n r) (/ (fact n) (fact r) (fact (- n r))))

(defun solve (a b v)
  (let ((ave (/ (loop for i below a sum (aref v i)) a))
        (num (count (aref v (1- a)) v))
        (tmp (loop for i from (1- a) downto 0 unless (= (aref v i) (aref v (1- a))) return (- a i 1))))
    (cons (float ave)
          (if (= (aref v 0) (aref v (1- a)))
            (loop for i from a to b when (<= i num) sum (ncr num i))
            (ncr num tmp)))))

(let* ((n (read))
       (a (read))
       (b (read))
       (v (sort (input n) #'>))
       (e (solve a b v)))
  (format t "~,6F~%~A~%" (car e) (cdr e)))


(let* ((a (read-char))
       (x (read-char))
       (b (read-char)))
  (if (char= a b) (princ "H") (princ "D")))

Score : 100 points

Problem Statement

Two deer, AtCoDeer and TopCoDeer, are playing a game called Honest or Dishonest.
In this game, an honest player always tells the truth, and an dishonest player always tell lies.
You are given two characters a and b as the input. Each of them is either H or D, and carries the following information:

If a=H, AtCoDeer is honest; if a=D, AtCoDeer is dishonest.
If b=H, AtCoDeer is saying that TopCoDeer is honest; if b=D, AtCoDeer is saying that TopCoDeer is dishonest.

Given this information, determine whether TopCoDeer is honest.

Constraints

a=H or a=D.

b=H or b=D.

Input

The input is given from Standard Input in the following format:

a b

Output

If TopCoDeer is honest, print H. If he is dishonest, print D.

Sample Input 1

H H

Sample Output 1

H

In this input, AtCoDeer is honest. Hence, as he says, TopCoDeer is honest.

Sample Input 2

D H

Sample Output 2

D

In this input, AtCoDeer is dishonest. Hence, contrary to what he says, TopCoDeer is dishonest.

Sample Input 3

D D

Sample Output 3

H


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           ;; enclose the form with VALUES to avoid being captured by LOOP macro
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
;;;
;;; Arithmetic operations with static modulus
;;;

;; FIXME: Currently MOD* and MOD+ doesn't apply MOD when the number of
;; parameters is one.
(defmacro define-mod-operations (divisor)
  `(progn
     (defun mod* (&rest args)
       (reduce (lambda (x y) (mod (* x y) ,divisor)) args))

     (sb-c:define-source-transform mod* (&rest args)
       (if (null args)
           1
           (reduce (lambda (x y) `(mod (* ,x ,y) ,',divisor)) args)))

     (defun mod- (&rest args)
       (reduce (lambda (x y) (mod (- x y) ,divisor)) args))

     (sb-c:define-source-transform mod- (&rest args)
       (if (null args)
           0
           (reduce (lambda (x y) `(mod (- ,x ,y) ,',divisor)) args)))

     (defun mod+ (&rest args)
       (reduce (lambda (x y) (mod (+ x y) ,divisor)) args))

     (sb-c:define-source-transform mod+ (&rest args)
       (if (null args)
           0
           (reduce (lambda (x y) `(mod (+ ,x ,y) ,',divisor)) args)))

     (define-modify-macro incfmod (delta)
       (lambda (x y) (mod (+ x y) ,divisor)))

     (define-modify-macro decfmod (delta)
       (lambda (x y) (mod (- x y) ,divisor)))

     (define-modify-macro mulfmod (multiplier)
       (lambda (x y) (mod (* x y) ,divisor)))))

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  "NOTE: cannot read -2^62"
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (setf minus t)))))))
      (declare ((integer 0 #.most-positive-fixnum) result))
      (loop
        (let* ((byte (%read-byte)))
          (if (<= 48 byte 57)
              (setq result (+ (- byte 48)
                              (* 10 (the (integer 0 #.(floor most-positive-fixnum 10))
                                         result))))
              (return (if minus (- result) result))))))))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(define-mod-operations +mod+)
(defun main ()
  (let* ((n (read))
         (k (read))
         (as (make-array n :element-type 'uint31))
         (dp (make-array (list (+ n 1) (+ k 1)) :element-type 'uint31))
         (dp2 (make-array (list n (+ k 1)) :element-type 'uint31))
         (powers (make-array 5000 :element-type 'uint31 :initial-element 1)))
    (declare (uint16 n k))
    (dotimes (i n)
      (setf (aref as i) (read-fixnum)))
    (dotimes (i (- (length powers) 1))
      (setf (aref powers (+ i 1))
            (mod (ash (aref powers i) 1) +mod+)))
    (dotimes (x (+ n 1))
      (dotimes (y (+ k 1))
        (setf (aref dp x y)
              (if (zerop x)
                  (if (zerop y) 1 0)
                  (mod+ (aref dp (- x 1) y)
                        (aref dp (- x 1) (max 0 (- y (aref as (- x 1))))))))))
    (dotimes (i n)
      (dotimes (y (+ k 1))
        (setf (aref dp2 i y)
              (if (zerop y)
                  (aref powers (- n 1))
                  (mod- (aref dp n y)
                        (aref dp2 i (max 0 (- y (aref as i)))))))))
    (println
     (loop for i below n
           count (= (aref dp n k)
                    (mod (ash (aref dp2 i k) 1) +mod+))))))

#-swank (main)

;;;
;;; Test and benchmark
;;;

#+swank
(defun io-equal (in-string out-string &key (function #'main) (test #'equal))
  "Passes IN-STRING to *STANDARD-INPUT*, executes FUNCTION, and returns true if
the string output to *STANDARD-OUTPUT* is equal to OUT-STRING."
  (labels ((ensure-last-lf (s)
             (if (eql (uiop:last-char s) #\Linefeed)
                 s
                 (uiop:strcat s uiop:+lf+))))
    (funcall test
             (ensure-last-lf out-string)
             (with-output-to-string (out)
               (let ((*standard-output* out))
                 (with-input-from-string (*standard-input* (ensure-last-lf in-string))
                   (funcall function)))))))

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    ;; (run-program "C:/Windows/System32/WindowsPowerShell/v1.0/powershell.exe" '("get-clipboard") :output out)
    (run-program "powershell.exe" '("-Command" "Get-Clipboard") :output out :search t)))

#+swank (defparameter *this-pathname* (uiop:current-lisp-file-pathname))
#+swank (defparameter *dat-pathname* (uiop:merge-pathnames* "test.dat" *this-pathname*))

#+swank
(defun run (&optional thing (out *standard-output*))
  "THING := null | string | symbol | pathname

null: run #'MAIN using the text on clipboard as input.
string: run #'MAIN using the string as input.
symbol: alias of FIVEAM:RUN!.
pathname: run #'MAIN using the text file as input."
  (let ((*standard-output* out))
    (etypecase thing
      (null
       (with-input-from-string (*standard-input* (delete #\Return (get-clipbrd)))
         (main)))
      (string
       (with-input-from-string (*standard-input* (delete #\Return thing))
         (main)))
      (symbol (5am:run! thing))
      (pathname
       (with-open-file (*standard-input* thing)
         (main))))))

#+swank
(defun gen-dat ()
  (uiop:with-output-file (out *dat-pathname* :if-exists :supersede)
    (format out "")))

#+swank
(defun bench (&optional (out (make-broadcast-stream)))
  (time (run *dat-pathname* out)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3 6
1 4 3
"
    "1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "5 400
3 1 4 1 5
"
    "5
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "6 20
10 4 3 10 25 2
"
    "3
")))

Score : 600 points

Problem Statement

AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i.
Because he loves big numbers, he calls a subset of the cards good when the sum of the numbers written on the cards in the subset, is K or greater.

Then, for each card i, he judges whether it is unnecessary or not, as follows:

If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary.

Otherwise, card i is NOT unnecessary.

Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.

Constraints

All input values are integers.

1≤N≤5000

1≤K≤5000

1≤a_i≤10^9 (1≤i≤N)

Partial Score

300 points will be awarded for passing the test set satisfying N,K≤400.

Input

The input is given from Standard Input in the following format:

N K
a_1 a_2 ... a_N

Output

Print the number of the unnecessary cards.

Sample Input 1

3 6
1 4 3

Sample Output 1

1

There are two good sets: {2,3} and {1,2,3}.

Card 1 is only contained in {1,2,3}, and this set without card 1, {2,3}, is also good. Thus, card 1 is unnecessary.

For card 2, a good set {2,3} without card 2, {3}, is not good. Thus, card 2 is NOT unnecessary.

Neither is card 3 for a similar reason, hence the answer is 1.

Sample Input 2

5 400
3 1 4 1 5

Sample Output 2

5

In this case, there is no good set. Therefore, all the cards are unnecessary.

Sample Input 3

6 20
10 4 3 10 25 2

Sample Output 3

3


(defun atari (X)
  (floor (1- (sqrt (1+ (* 8 X)))) 2) )

(defun solve (X &optional (Y (1- (atari X))))
  (if (<= X (/ (* Y (1+ Y)) 2)) Y
    (solve X (1+ Y)) ))

(princ (solve(read)))

Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0.
During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right.
That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i.
The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible.
Find the earliest possible time to reach coordinate X.

Constraints

X is an integer.

1≤X≤10^9

Input

The input is given from Standard Input in the following format:

X

Output

Print the earliest possible time for the kangaroo to reach coordinate X.

Sample Input 1

6

Sample Output 1

3

The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.

Sample Input 2

2

Sample Output 2

2

He can reach his nest at time 2 by staying at his position during the first second, and jumping to the right at the next second.

Sample Input 3

11

Sample Output 3

5


(locally (declare (optimize (speed 3) (safety 0)))
  (defun split-with (string &key (delimiterp #'(lambda (c) (char= c #\Space))))
    (loop :for beg = (position-if-not delimiterp string)
          :then (position-if-not delimiterp string :start (1+ end))
          :for end = (and beg (position-if delimiterp string :start beg))
          :when beg :collect (subseq string beg end)
          :while end))

  (defun count-bus (n c k t-list)
    (declare (ignore n))
    (declare (integer c k))
    (let ((t0 (aref t-list 0))
          (ptr 0)
          (count 0))
      (loop
        :while (< ptr (- n 1))
        :do (loop
              :for i :downfrom (if (< (length t-list) c)
                                   (- (length t-list) 1)
                                   (- c 1))
              :when (<= (aref t-list (+ i ptr)) (+ t0 k))
              :do (progn (incf count)
                         (setf ptr (+ i ptr 1))
                         (unless (zerop (length t-list))
                           (setf t0 (aref t-list ptr)))
                         (loop-finish))))
      count))

  (let* ((fl (split-with (read-line t nil nil)))
         (n (parse-integer (car fl)))
         (c (parse-integer (cadr fl)))
         (k (parse-integer (caddr fl)))
         (t-list (coerce (sort (loop :for x = (read-line t nil nil)
                                     :repeat n
                                     :while x
                                     :collect (parse-integer x))
                               #'<)
                         'vector)))
    (format t "~d~%" (count-bus n c k t-list))))

Score : 300 points

Problem Statement

Every day, N passengers arrive at Takahashi Airport.
The i-th passenger arrives at time T_i.

Every passenger arrived at Takahashi airport travels to the city by bus. Each bus can accommodate up to C passengers.
Naturally, a passenger cannot take a bus that departs earlier than the airplane arrives at the airport.
Also, a passenger will get angry if he/she is still unable to take a bus K units of time after the arrival of the airplane.
For that reason, it is necessary to arrange buses so that the i-th passenger can take a bus departing at time between T_i and T_i + K (inclusive).

When setting the departure times for buses under this condition, find the minimum required number of buses.
Here, the departure time for each bus does not need to be an integer, and there may be multiple buses that depart at the same time.

Constraints

2 \leq N \leq 100000

1 \leq C \leq 10^9

1 \leq K \leq 10^9

1 \leq T_i \leq 10^9

C, K and T_i are integers.

Input

The input is given from Standard Input in the following format:

N C K
T_1
T_2
:
T_N

Output

Print the minimum required number of buses.

Sample Input 1

5 3 5
1
2
3
6
12

Sample Output 1

3

For example, the following three buses are enough:

A bus departing at time 4.5, that carries the passengers arriving at time 2 and 3.

A bus departing at time 6, that carries the passengers arriving at time 1 and 6.

A bus departing at time 12, that carries the passenger arriving at time 12.

Sample Input 2

6 3 3
7
6
2
8
10
6

Sample Output 2

3


(declaim (optimize (speed 3) (debug 0) (safety 0)))
(defvar n (read))
(defvar a (make-array n))
(loop for i from 0 below n
      do (setf (aref a i) (read)))
(sort a #'>)
(defun solve ()
  (loop for i from 1 below n
        for able = 0
        do (loop for j from (1- n) downto 0
                 initially (setq able (aref a i))
                 unless (= j i)
                 if (> (aref a j) (* able 2))
                 do (return-from solve i)
                 else do (setq able (+ able (aref a j)))
                 finally (setq able 0)))
  n)
(format t "~A~%" (solve))

Score : 400 points

Problem Statement

Snuke found N strange creatures.
Each creature has a fixed color and size. The color and size of the i-th creature are represented by i and A_i, respectively.

Every creature can absorb another creature whose size is at most twice the size of itself.
When a creature of size A and color B absorbs another creature of size C and color D (C \leq 2 \times A), they will merge into one creature of size A+C and color B.
Here, depending on the sizes of two creatures, it is possible that both of them can absorb the other.

Snuke has been watching these creatures merge over and over and ultimately become one creature.
Find the number of the possible colors of this creature.

Constraints

2 \leq N \leq 100000

1 \leq A_i \leq 10^9

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

Print the number of the possible colors of the last remaining creature after the N creatures repeatedly merge and ultimately become one creature.

Sample Input 1

3
3 1 4

Sample Output 1

2

The possible colors of the last remaining creature are colors 1 and 3.
For example, when the creature of color 3 absorbs the creature of color 2, then the creature of color 1 absorbs the creature of color 3, the color of the last remaining creature will be color 1.

Sample Input 2

5
1 1 1 1 1

Sample Output 2

5

There may be multiple creatures of the same size.

Sample Input 3

6
40 1 30 2 7 20

Sample Output 3

4


(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           ;; enclose the form with VALUES to avoid being captured by LOOP macro
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(macrolet ((def (name fname)
             `(define-modify-macro ,name (new-value) ,fname)))
  (def minf min)
  (def maxf max)
  (def mulf *)
  (def divf /)
  (def iorf logior)
  (def xorf logxor)
  (def andf logand))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun main ()
  (let* ((n (read))
         feasible
         (plan (make-array (list n n) :element-type 'bit :initial-element 0))
         (rows (make-array n :element-type 'uint32 :initial-element 0))
         (cols (make-array n :element-type 'uint32 :initial-element 0)))
    (dotimes (i n)
      (let ((line (read-line)))
        (dotimes (j n)
          (when (char= #\# (aref line j))
            (setf (aref plan i j) 1)
            (setq feasible t)
            (incf (aref rows i))
            (incf (aref cols j))))))
    (unless feasible
      (println -1)
      (return-from main))
    (println
     (+ (- n (count n cols))
        (loop for row below n
              minimize (+ (loop for j below n
                                count (zerop (aref plan row j)))
                          (if (loop for i below n
                                    always (zerop (aref plan i row)))
                              1
                              0)))))))

#-swank (main)

;;;
;;; Test and benchmark
;;;

#+swank
(defun io-equal (in-string out-string &key (function #'main) (test #'equal))
  "Passes IN-STRING to *STANDARD-INPUT*, executes FUNCTION, and returns true if
the string output to *STANDARD-OUTPUT* is equal to OUT-STRING."
  (labels ((ensure-last-lf (s)
             (if (eql (uiop:last-char s) #\Linefeed)
                 s
                 (uiop:strcat s uiop:+lf+))))
    (funcall test
             (ensure-last-lf out-string)
             (with-output-to-string (out)
               (let ((*standard-output* out))
                 (with-input-from-string (*standard-input* (ensure-last-lf in-string))
                   (funcall function)))))))

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    (run-program "powershell.exe" '("-Command" "Get-Clipboard") :output out :search t)))

#+swank (defparameter *this-pathname* (uiop:current-lisp-file-pathname))
#+swank (defparameter *dat-pathname* (uiop:merge-pathnames* "test.dat" *this-pathname*))

#+swank
(defun run (&optional thing (out *standard-output*))
  "THING := null | string | symbol | pathname

null: run #'MAIN using the text on clipboard as input.
string: run #'MAIN using the string as input.
symbol: alias of FIVEAM:RUN!.
pathname: run #'MAIN using the text file as input."
  (let ((*standard-output* out))
    (etypecase thing
      (null
       (with-input-from-string (*standard-input* (delete #\Return (get-clipbrd)))
         (main)))
      (string
       (with-input-from-string (*standard-input* (delete #\Return thing))
         (main)))
      (symbol (5am:run! thing))
      (pathname
       (with-open-file (*standard-input* thing)
         (main))))))

#+swank
(defun gen-dat ()
  (uiop:with-output-file (out *dat-pathname* :if-exists :supersede)
    (format out "")))

#+swank
(defun bench (&optional (out (make-broadcast-stream)))
  (time (run *dat-pathname* out)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
#.
.#
"
    "3
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
..
..
"
    "-1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
##
##
"
    "0
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
.#.
###
.#.
"
    "2
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
...
.#.
...
"
    "5
")))

Score : 1300 points

Problem Statement

There is a square-shaped grid with N vertical rows and N horizontal columns.
We will denote the square at the i-th row from the top and the j-th column from the left as (i,\ j).

Initially, each square is either white or black.
The initial color of the grid is given to you as characters a_{ij}, arranged in a square shape.
If the square (i,\ j) is white, a_{ij} is .. If it is black, a_{ij} is #.

You are developing a robot that repaints the grid.
It can repeatedly perform the following operation:

Select two integers i, j (1 ≤ i,\ j ≤ N). Memorize the colors of the squares (i,\ 1), (i,\ 2), ..., (i,\ N) as c_1, c_2, ..., c_N, respectively. Then, repaint the squares (1,\ j), (2,\ j), ..., (N,\ j) with the colors c_1, c_2, ..., c_N, respectively.

Your objective is to turn all the squares black.
Determine whether it is possible, and find the minimum necessary number of operations to achieve it if the answer is positive.

Constraints

2 ≤ N ≤ 500

a_{ij} is either . or #.

Partial Score

In a test set worth 300 points, N ≤ 3.

Input

The input is given from Standard Input in the following format:

N
a_{11}...a_{1N}
:
a_{N1}...a_{NN}

Output

If it is possible to turn all the squares black, print the minimum necessary number of operations to achieve the objective.
If it is impossible, print -1 instead.

Sample Input 1

2
#.
.#

Sample Output 1

3

For example, perform the operation as follows:

Select i = 1, j = 2.

Select i = 1, j = 1.

Select i = 1, j = 2.

The transition of the colors of the squares is shown in the figure below:

Sample Input 2

2
..
..

Sample Output 2

-1

Sample Input 3

2
##
##

Sample Output 3

0

Sample Input 4

3
.#.
###
.#.

Sample Output 4

2

Sample Input 5

3
...
.#.
...

Sample Output 5

5


(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           ;; enclose the form with VALUES to avoid being captured by LOOP macro
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(macrolet ((def (name fname)
             `(define-modify-macro ,name (new-value) ,fname)))
  (def minf min)
  (def maxf max)
  (def mulf *)
  (def divf /)
  (def iorf logior)
  (def xorf logxor)
  (def andf logand))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun main ()
  (let* ((n (read))
         (diagonal 0)
         (safe 0)
         (plan (make-array (list n n) :element-type 'bit :initial-element 0))
         (rows (make-array n :element-type 'uint32 :initial-element 0))
         (cols (make-array n :element-type 'uint32 :initial-element 0)))
    (dotimes (i n)
      (let ((line (read-line)))
        (dotimes (j n)
          (when (char= #\# (aref line j))
            (setf (aref plan i j) 1)
            (if (= i j)
                (incf diagonal)
                (incf safe))
            (incf (aref rows i))
            (incf (aref cols j))))))
    (when (= 0 diagonal safe)
      (println -1)
      (return-from main))
    (when (find n rows)
      (println (- n (count n cols)))
      (return-from main))
    (when (find n cols)
      (println (- n (count n rows)))
      (return-from main))
    (let ((base (if (zerop safe)
                    #xffffffff
                    (+ 1 (min (reduce #'max rows :key (lambda (x) (- n x)))
                              (reduce #'max cols :key (lambda (x) (- n x))))))))
      #>base
      (dotimes (i n)
        (dotimes (j n)
          (when (= 1 (aref plan i j))
            (minf base (- n (aref rows j)))
            (minf base (- n (aref cols i)))
            (when (= i j)
              (minf base
                    (+ 1 (min (loop for row below n
                                    when (= 1 (aref plan row i))
                                    minimize (- n (aref rows row)))
                              (loop for col below n
                                    when (= 1 (aref plan j col))
                                    minimize (- n (aref cols col))))))))))
      (assert (< base #xffffffff))
      (println (+ base n)))))

#-swank (main)

;;;
;;; Test and benchmark
;;;

#+swank
(defun io-equal (in-string out-string &key (function #'main) (test #'equal))
  "Passes IN-STRING to *STANDARD-INPUT*, executes FUNCTION, and returns true if
the string output to *STANDARD-OUTPUT* is equal to OUT-STRING."
  (labels ((ensure-last-lf (s)
             (if (eql (uiop:last-char s) #\Linefeed)
                 s
                 (uiop:strcat s uiop:+lf+))))
    (funcall test
             (ensure-last-lf out-string)
             (with-output-to-string (out)
               (let ((*standard-output* out))
                 (with-input-from-string (*standard-input* (ensure-last-lf in-string))
                   (funcall function)))))))

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    (run-program "powershell.exe" '("-Command" "Get-Clipboard") :output out :search t)))

#+swank (defparameter *this-pathname* (uiop:current-lisp-file-pathname))
#+swank (defparameter *dat-pathname* (uiop:merge-pathnames* "test.dat" *this-pathname*))

#+swank
(defun run (&optional thing (out *standard-output*))
  "THING := null | string | symbol | pathname

null: run #'MAIN using the text on clipboard as input.
string: run #'MAIN using the string as input.
symbol: alias of FIVEAM:RUN!.
pathname: run #'MAIN using the text file as input."
  (let ((*standard-output* out))
    (etypecase thing
      (null
       (with-input-from-string (*standard-input* (delete #\Return (get-clipbrd)))
         (main)))
      (string
       (with-input-from-string (*standard-input* (delete #\Return thing))
         (main)))
      (symbol (5am:run! thing))
      (pathname
       (with-open-file (*standard-input* thing)
         (main))))))

#+swank
(defun gen-dat ()
  (uiop:with-output-file (out *dat-pathname* :if-exists :supersede)
    (format out "")))

#+swank
(defun bench (&optional (out (make-broadcast-stream)))
  (time (run *dat-pathname* out)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
#.
.#
"
    "3
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
..
..
"
    "-1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
##
##
"
    "0
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
.#.
###
.#.
"
    "2
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
...
.#.
...
"
    "5
")))


Score : 100 points

Problem Statement

Snuke has a favorite restaurant.

The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.

So far, Snuke has ordered N meals at the restaurant.
Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen.
Find x-y.

Constraints

1 ≤ N ≤ 100

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

20

Sample Output 1

15800

So far, Snuke has paid 16000 yen, and the restaurant has paid back 200 yen. Thus, the answer is 15800.

Sample Input 2

60

Sample Output 2

47200

Snuke has paid 48000 yen for 60 meals, and the restaurant has paid back 800 yen.


(let ((n (read))
      (s 1))

  (loop for i from 1 to n
        do (setq s (mod (* s i) (+ (expt 10 9) 7))))

  (format t "~A~%" s))

Score : 200 points

Problem Statement

Snuke loves working out. He is now exercising N times.

Before he starts exercising, his power is 1. After he exercises for the i-th time, his power gets multiplied by i.

Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

Constraints

1 ≤ N ≤ 10^{5}

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer modulo 10^{9}+7.

Sample Input 1

3

Sample Output 1

6

After Snuke exercises for the first time, his power gets multiplied by 1 and becomes 1.

After Snuke exercises for the second time, his power gets multiplied by 2 and becomes 2.

After Snuke exercises for the third time, his power gets multiplied by 3 and becomes 6.

Sample Input 2

10

Sample Output 2

3628800

Sample Input 3

100000

Sample Output 3

457992974

Print the answer modulo 10^{9}+7.


(setq s 1)
(dotimes(i(read))(setq s(mod(* s(1+ i))1000000007)))
(princ s)


(defun split (str delim)
  (let ((res (make-array 1 :element-type 'string
                           :fill-pointer 0
                           :adjustable t)))
    (loop for i from 0 below (length str)
          with start = 0
          when (eq (char str i) delim)
          do (vector-push-extend (subseq str start i) res)
             (setf start (1+ i))
          finally (let ((tail (subseq str start)))
                    (when tail
                      (vector-push-extend tail res))))
    res))

(let* ((line (map 'vector #'parse-integer (split (read-line) #\Space)))
       (n (aref line 0))
       (m (truncate (aref line 1) 2))
       (res 0)
       (nmin (min n m)))
  (incf res nmin)
  (decf n nmin)
  (decf m nmin)
  (incf res (truncate m 2))
  (format t "~A~%" res))

Score : 300 points

Problem Statement

Snuke loves puzzles.

Today, he is working on a puzzle using S- and c-shaped pieces.
In this puzzle, you can combine two c-shaped pieces into one S-shaped piece, as shown in the figure below:

Snuke decided to create as many Scc groups as possible by putting together one S-shaped piece and two c-shaped pieces.

Find the maximum number of Scc groups that can be created when Snuke has N S-shaped pieces and M c-shaped pieces.

Constraints

1 ≤ N,M ≤ 10^{12}

Input

The input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

1 6

Sample Output 1

2

Two Scc groups can be created as follows:

Combine two c-shaped pieces into one S-shaped piece

Create two Scc groups, each from one S-shaped piece and two c-shaped pieces

Sample Input 2

12345 678901

Sample Output 2

175897


(defun calc (A1 AN S)
  (loop for si in S
        with a = A1
        and  b = AN
        collect (if a #\S #\W)
        do
        (shiftf b a (eq (eq a b) si))
        finally (unless (eq A1 a) (return nil)) ))

(defun solve (l)
  (concatenate 'string
               (or (calc t t l)
                   (calc t nil l)
                   (calc nil t l)
                   (calc nil nil l)
                   "-1")))
(read)
(princ (solve
        (mapcar (lambda (x) (eq x #\o))
                (concatenate 'list
                             (read-line)) )))

Score : 500 points

Problem Statement

Snuke, who loves animals, built a zoo.

There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle.
The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1.

There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies.

Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is o, the animal said that the two neighboring animals are of the same species, and if s_i is x, the animal said that the two neighboring animals are of different species.

More formally, a sheep answered o if the two neighboring animals are both sheep or both wolves, and answered x otherwise.
Similarly, a wolf answered x if the two neighboring animals are both sheep or both wolves, and answered o otherwise.

Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print -1.

Constraints

3 ≤ N ≤ 10^{5}

s is a string of length N consisting of o and x.

Input

The input is given from Standard Input in the following format:

N
s

Output

If there does not exist an valid assignment that is consistent with s, print -1.
Otherwise, print an string t in the following format. The output is considered correct if the assignment described by t is consistent with s.

t is a string of length N consisting of S and W.

If t_i is S, it indicates that the animal numbered i is a sheep. If t_i is W, it indicates that the animal numbered i is a wolf.

Sample Input 1

6
ooxoox

Sample Output 1

SSSWWS

For example, if the animals numbered 1, 2, 3, 4, 5 and 6 are respectively a sheep, sheep, sheep, wolf, wolf, and sheep, it is consistent with their responses. Besides, there is another valid assignment of species: a wolf, sheep, wolf, sheep, wolf and wolf.

Let us remind you: if the neiboring animals are of the same species, a sheep answers o and a wolf answers x. If the neiboring animals are of different species, a sheep answers x and a wolf answers o.

Sample Input 2

3
oox

Sample Output 2

-1

Print -1 if there is no valid assignment of species.

Sample Input 3

10
oxooxoxoox

Sample Output 3

SSWWSSSWWS


Score : 100 points

Problem Statement

Alice and Bob are playing One Card Poker.

One Card Poker is a two-player game using playing cards.

Each card in this game shows an integer between 1 and 13, inclusive.

The strength of a card is determined by the number written on it, as follows:

Weak 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10 < 11 < 12 < 13 < 1 Strong

One Card Poker is played as follows:

Each player picks one card from the deck. The chosen card becomes the player's hand.

The players reveal their hands to each other. The player with the stronger card wins the game.

If their cards are equally strong, the game is drawn.

You are watching Alice and Bob playing the game, and can see their hands.

The number written on Alice's card is A, and the number written on Bob's card is B.

Write a program to determine the outcome of the game.

Constraints

1≦A≦13

1≦B≦13

A and B are integers.

Input

The input is given from Standard Input in the following format:

A B

Output

Print Alice if Alice will win. Print Bob if Bob will win. Print Draw if the game will be drawn.

Sample Input 1

8 6

Sample Output 1

Alice

8 is written on Alice's card, and 6 is written on Bob's card.
Alice has the stronger card, and thus the output should be Alice.

Sample Input 2

1 1

Sample Output 2

Draw

Since their cards have the same number, the game will be drawn.

Sample Input 3

13 1

Sample Output 3

Bob


(defun split (str delim)
  (let ((res (make-array 1 :element-type 'string
                           :fill-pointer 0
                           :adjustable t)))
    (loop for i from 0 below (length str)
          with start = 0
          when (eq (char str i) delim)
          do (vector-push-extend (subseq str start i) res)
             (setf start (1+ i))
          finally (let ((tail (subseq str start)))
                    (when tail
                      (vector-push-extend tail res))))
    res))

(let* ((l (map 'vector 'parse-integer (split (read-line) #\ )))
       (a (aref l 0))
       (b (aref l 1)))
  (when (= a 1) (setf a 14))
  (when (= b 1) (setf b 14))
  (if (= a b)
      (format t "Draw~%")
      (if (> a b)
          (format t "Alice~%")
          (format t "Bob~%"))))


(defun list-slice (lst x1 x2 y1 y2)
  (mapcar (lambda (k) (subseq k x1 x2)) (subseq lst y1 y2)))
(let* ((n (read))
       (m (read))
       (lst-a (loop :repeat n :collect (concatenate 'list (read-line))))
       (lst-b (loop :repeat m :collect (concatenate 'list (read-line)))))
  (if (loop :for x :from 0 :upto (- n m)
            :never(loop :for y :from 0 :upto (- n m) :never (equal lst-b (list-slice lst-a x (+ x m) y (+ y m)))))
      (format t "Yes~%")
      (format t "No~%")))

Score : 200 points

Problem Statement

You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.

A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.

Also, the given images are binary images, and the color of each pixel is either white or black.

In the input, every pixel is represented by a character: . corresponds to a white pixel, and # corresponds to a black pixel.

The image A is given as N strings A_1,...,A_N.

The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).

Similarly, the template image B is given as M strings B_1,...,B_M.

The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).

Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Constraints

1≦M≦N≦50

A_i is a string of length N consisting of # and ..

B_i is a string of length M consisting of # and ..

Input

The input is given from Standard Input in the following format:

N M
A_1
A_2
:
A_N
B_1
B_2
:
B_M

Output

Print Yes if the template image B is contained in the image A. Print No otherwise.

Sample Input 1

3 2
#.#
.#.
#.#
#.
.#

Sample Output 1

Yes

The template image B is identical to the upper-left 2 × 2 subimage and the lower-right 2 × 2 subimage of A. Thus, the output should be Yes.

Sample Input 2

4 1
....
....
....
....
#

Sample Output 2

No

The template image B, composed of a black pixel, is not contained in the image A composed of white pixels.