#-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))))
(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)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  "NOTE: cannot read -2^62"
  (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 #\-))
                                  (setq 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))))))))

;; PAY ATTENTION TO THE STACK SIZE!
(declaim (inline bipartite-p)
         (ftype (function * (values (or null simple-bit-vector) &optional)) bipartite-p))
(defun bipartite-p (graph)
  "Checks if GRAPH is bipartite and returns the vector of colorings if so,
otherwise returns NIL.

GRAPH := vector of adjacency lists"
  (declare (vector graph))
  (let* ((n (length graph))
         (visited (make-array n :element-type 'bit :initial-element 0))
         (colors (make-array n :element-type 'bit :initial-element 0)))
    (labels ((dfs (vertex color)
               (cond ((zerop (aref visited vertex))
                      (setf (aref visited vertex) 1
                            (aref colors vertex) color)
                      (if (= color 1)
                          (dolist (neighbor (aref graph vertex))
                            (dfs neighbor 0))
                          (dolist (neighbor (aref graph vertex))
                            (dfs neighbor 1))))
                     ((/= color (aref colors vertex))
                      (return-from bipartite-p nil)))))
      (dotimes (i n colors)
        (when (zerop (aref visited i))
          (dfs i 1))))))

(in-package :cl-user)

(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))
         (m (read))
         (graph (make-array n :element-type 'list :initial-element nil)))
    (declare (uint31 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 ((colors (bipartite-p graph)))
      (println
       (if colors
           (let* ((size1 (count 0 colors))
                  (size2 (- n size1)))
             (- (* size1 size2) m))
           (- (ash (* n (- n 1)) -1) m))))))

#-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)
    #+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* 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 "6 5
1 2
2 3
3 4
4 5
5 6
"
    "4
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "5 5
1 2
2 3
3 1
5 4
5 1
"
    "5
")))

Score : 500 points

Problem Statement

Rng has a connected undirected graph with N vertices.
Currently, there are M edges in the graph, and the i-th edge connects Vertices A_i and B_i.

Rng will add new edges to the graph by repeating the following operation:

Operation: Choose u and v (u \neq v) such that Vertex v can be reached by traversing exactly three edges from Vertex u, and add an edge connecting Vertices u and v. It is not allowed to add an edge if there is already an edge connecting Vertices u and v.

Find the maximum possible number of edges that can be added.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i,B_i \leq N

The graph has no self-loops or multiple edges.

The graph is connected.

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 the maximum possible number of edges that can be added.

Sample Input 1

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

Sample Output 1

4

If we add edges as shown below, four edges can be added, and no more.

Sample Input 2

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

Sample Output 2

5

Five edges can be added, for example, as follows:

Add an edge connecting Vertex 5 and Vertex 3.

Add an edge connecting Vertex 5 and Vertex 2.

Add an edge connecting Vertex 4 and Vertex 1.

Add an edge connecting Vertex 4 and Vertex 2.

Add an edge connecting Vertex 4 and Vertex 3.


;; -*- 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
(declaim (inline relative-error-p))
(defun relative-error<= (x y threshold)
  "Returns true if the relative error between X and Y is equal to or smaller
than THRESHOLD: i.e. the relative errors of any numbers in the interval [X,
Y] (or [Y, X]) are equal to or smaller than THRESHOLD when the true value is in
the same interval."
  (and (not (zerop x))
       (not (zerop y))
       (<= (abs (/ (- x y) y)) threshold)
       (<= (abs (/ (- x y) x)) threshold)))

;;;
;;; ARRAY-ELEMENT-TYPE is not constant-folded on SBCL version earlier than
;;; 1.5.0. See
;;; https://github.com/sbcl/sbcl/commit/9f0d12e7ab961828931d01c0b2a76a5885ad35d2
;;;

(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-c:deftransform array-element-type ((array))
    (let ((type (sb-c::lvar-type array)))
      (flet ((element-type (type)
               (and (sb-c::array-type-p type)
                    (sb-int:neq (sb-kernel::array-type-specialized-element-type type) sb-kernel:*wild-type*)
                    (sb-kernel:type-specifier (sb-kernel::array-type-specialized-element-type type)))))
        (cond ((let ((type (element-type type)))
                 (and type
                      `',type)))
              ((sb-kernel:union-type-p type)
               (let (result)
                 (loop for type in (sb-kernel:union-type-types type)
                       for et = (element-type type)
                       unless (and et
                                   (if result
                                       (equal result et)
                                       (setf result et)))
                       do (sb-c::give-up-ir1-transform))
                 `',result))
              ((sb-kernel:intersection-type-p type)
               (loop for type in (sb-kernel:intersection-type-types type)
                     for et = (element-type type)
                     when et
                     return `',et
                     finally (sb-c::give-up-ir1-transform)))
              (t
               (sb-c::give-up-ir1-transform)))))))

;;;
;;; Compute inversion number by merge sort
;;;

(declaim (inline %merge-count))
(defun %merge-count (l mid r source-vec dest-vec predicate)
  (declare ((integer 0 #.array-total-size-limit) l mid r)
           (function predicate))
  (loop with count of-type (integer 0 #.most-positive-fixnum) = 0
        with i = l
        with j = mid
        for idx from l
        when (= i mid)
        do (loop for j from j below r
                 for idx from idx
                 do (setf (aref dest-vec idx)
                          (aref source-vec j))
                 finally (return-from %merge-count count))
        when (= j r)
        do (loop for i from i below mid
                 for idx from idx
                 do (setf (aref dest-vec idx)
                          (aref source-vec i))
                 finally (return-from %merge-count count))
        do (if (funcall predicate
                        (aref source-vec j)
                        (aref source-vec i))
               (setf (aref dest-vec idx) (aref source-vec j)
                     j (1+ j)
                     count (+ count (- mid i)))
               (setf (aref dest-vec idx) (aref source-vec i)
                     i (1+ i)))))

(defmacro with-fixnum+ (form)
  (let ((fixnum+ '(integer 0 #.most-positive-fixnum)))
    `(the ,fixnum+
          ,(reduce (lambda (f1 f2)`(,(car form)
                                   (the ,fixnum+ ,f1)
                                   (the ,fixnum+ ,f2)))
	           (cdr form)))))

(declaim (inline %calc-by-insertion-sort!))
(defun %calc-by-insertion-sort! (vec predicate l r)
  (declare (function predicate)
           ((integer 0 #.array-total-size-limit) l r))
  (loop with inv-count of-type (integer 0 #.most-positive-fixnum) = 0
        for end from (+ l 1) below r
        do (loop for i from end above l
                 while (funcall predicate (aref vec i) (aref vec (- i 1)))
                 do (rotatef (aref vec (- i 1)) (aref vec i))
                    (incf inv-count))
        finally (return inv-count)))

;; NOTE: This function is slow on SBCL version earlier than 1.5.0 as
;; constant-folding of ARRAY-ELEMENT-TYPE doesn't work. Use
;; array-element-type.lisp if necessary.
(declaim (inline calc-inversion-number!))
(defun calc-inversion-number! (vector predicate &key (start 0) end buffer)
  "Calculates the inversion number of VECTOR w.r.t. the strict order
PREDICATE. This function sorts VECTOR as a side effect."
  (declare ((simple-array double-float (*)) vector buffer)
           (function predicate))
  (let ((end (or end (length vector))))
    (declare ((integer 0 #.array-total-size-limit) start end))
    (assert (<= start end))
    (labels
        ((recurse (l r merge-to-vec1-p)
           (declare (optimize (safety 0))
                    ((integer 0 #.array-total-size-limit) l r))
           (cond ((and (<= (- r l) 24) merge-to-vec1-p)
                  (%calc-by-insertion-sort! vector predicate l r))
                 (t
                  (let ((mid (floor (+ l r) 2)))
                    (with-fixnum+
                        (+ (recurse l mid (not merge-to-vec1-p))
                           (recurse mid r (not merge-to-vec1-p))
                           (if merge-to-vec1-p
                               (%merge-count l mid r buffer vector predicate)
                               (%merge-count l mid r vector buffer predicate)))))))))
      (recurse start end t))))

(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 solve (n as bs cs)
  (declare #.OPT
           ((simple-array int32 (*)) as bs cs)
           (uint16 n))
  (let ((lefts (make-array n :element-type 'double-float))
        (rights (make-array n :element-type 'double-float))
        (median (ceiling (floor (* n (- n 1)) 2) 2))
        (buffer (make-array n :element-type 'double-float)))
    (dotimes (i n)
      (let ((y (/ (+ (aref cs i) (* 1d9 (aref as i)))
                  (aref bs i))))
        (setf (aref lefts i) y)))
    (let ((ords (make-array n :element-type 'uint32)))
      (dotimes (i n) (setf (aref ords i) i))
      (setf ords (sort ords
                       (lambda (i j)
                         (< (aref lefts i) (aref lefts j)))))
      (sb-int:named-let bisect ((ng -1d9) (ok 1d9))
        (declare (double-float ng ok))
        (if (or (< (- ok ng) 1d-10)
                (relative-error<= ng ok 1d-10))
            ok
            (let ((mid (* 0.5d0 (+ ng ok))))
              (dotimes (i n)
                (let* ((ord (aref ords i))
                       (y (/ (- (aref cs ord) (* mid (aref as ord)))
                             (aref bs ord))))
                  (setf (aref rights i) y)))
              (let ((invs (calc-inversion-number! rights #'< :buffer buffer)))
                (if (>= invs median)
                    (bisect ng mid)
                    (bisect mid ok)))))))))

(defun main ()
  (let* ((n (read))
         (as (make-array n :element-type 'int32))
         (bs (make-array n :element-type 'int32))
         (cs (make-array n :element-type 'int32)))
    (dotimes (i n)
      (setf (aref as i) (read-fixnum)
            (aref bs i) (read-fixnum)
            (aref cs i) (read-fixnum)))
    (let ((x (solve n as bs cs))
          (y (solve n bs as cs)))
      (format t "~F ~F~%" x y))))

#-swank (main)

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

Score : 800 points

Problem Statement

There are N lines in the xy-plane. The i-th line is represented by A_ix+B_iy=C_i.
Any two lines among the N+2 lines, the above N lines plus the x-axis and y-axis, cross each other at exactly one point.

For each pair 1 \leq i < j \leq N, there is a car at the cross point of the i-th and j-th lines.
Even where three or more lines intersect at a point, a car is individually placed for each pair of lines.
That is, there will be k(k-1)/2 cars placed at the intersection of k lines.

Those cars are already very old, and can only be moved parallel to the x-axis or y-axis.

Takahashi will hold an exhibition of antique cars at a place on the xy-plane.
In order to avoid damaging the half-broken cars too much, he will select the place of the exhibition so that the total distance covered will be minimized when all the cars are moved to the place.
If such a place is not uniquely determined, among the places that satisfy the condition above, the place with the minimum x-coordinate will be selected.
If the place is still not uniquely determined, among the places that satisfy the two conditions above, the place with the minimum y-coordinate will be selected.

Find the place of the exhibition that will be selected.

Constraints

2 \leq N \leq 4 × 10^4

1 \leq |A_i|,|B_i| \leq 10^4(1 \leq i \leq N)

0 \leq |C_i| \leq 10^4(1 \leq i \leq N)

No two given lines are parallel.

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
A_1 B_1 C_1
:
A_N B_N C_N

Outputs

Print the x-coordinate and y-coordinate of the place of the exhibition that will be selected, in this order, with a space in between. The output will be judged as correct when the absolute or relative error is at most 10^{-9}.

Sample Input 1

3
1 1 1
2 -1 2
-1 2 2

Sample Output 1

1.000000000000000 1.000000000000000

There is a car at each place shown by a blue circle in the figure. The place to be selected is shown by a purple circle.

Sample Input 2

4
1 1 2
1 -1 0
3 -1 -2
1 -3 4

Sample Output 2

-1.000000000000000 -1.000000000000000

Sample Input 3

7
1 7 8
-2 4 9
3 -8 -5
9 2 -14
6 7 5
-8 -9 3
3 8 10

Sample Output 3

-1.722222222222222 1.325000000000000

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.


Score : 100 points

Problem Statement

Snuke prepared 6 problems for a upcoming programming contest.
For each of those problems, Rng judged whether it can be used in the contest or not.

You are given a string S of length 6.
If the i-th character of s is 1, it means that the i-th problem prepared by Snuke is accepted to be used; 0 means that the problem is not accepted.

How many problems prepared by Snuke are accepted to be used in the contest?

Constraints

The length of S is 6.

S consists of 0 and 1.

Inputs

Input is given from Standard Input in the following format:

S

Outputs

Print the number of problems prepared by Snuke that are accepted to be used in the contest.

Sample Input 1

111100

Sample Output 1

4

The first, second, third and fourth problems are accepted, for a total of four.

Sample Input 2

001001

Sample Output 2

2

Sample Input 3

000000

Sample Output 3

0


(setq s (read-line))

(princ (if (and (<= 4 (length s)) (string= (subseq s 0 4) "YAKI"))
		 "Yes"
		 "No"))

Score : 100 points

Problem Statement

Ringo is giving a present to Snuke.

Ringo has found out that Snuke loves yakiniku (a Japanese term meaning grilled meat. yaki: grilled, niku: meat). He supposes that Snuke likes grilled things starting with YAKI in Japanese, and does not like other things.

You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with YAKI.

Constraints

1 \leq |S| \leq 10

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S starts with YAKI, print Yes; otherwise, print No.

Sample Input 1

YAKINIKU

Sample Output 1

Yes

YAKINIKU starts with YAKI.

Sample Input 2

TAKOYAKI

Sample Output 2

No

TAKOYAKI (a Japanese snack. tako: octopus) does not start with YAKI.

Sample Input 3

YAK

Sample Output 3

No


(format t "~A~%" (if (string= "YAKI" (concatenate 'string (read-line) "    ")
                              :end2 4)
                     "Yes"
                     "No"))


(defun f ()
  (let ((N (read)) (M (read)) (K (read)))
    (defun fLIP? (N M K)
      (defun iter (i j)
	  (if (= i (+ N 1))
	      nil
	      (if (= (+ M 1) j)
		  (iter (+ i 1) 0)
		  (if (= (+ (* i j) (* (- N i) (- M j)))
			 K)
		      t
		      (iter i (+ j 1))))))
      (iter 0 0))
    (if (fLIP? N M K)
	(format t "~A~%" "Yes")
	(format t "~A~%" "No"))))

(f)

Score : 200 points

Problem Statement

We have a grid with N rows and M columns of squares. Initially, all the squares are white.

There is a button attached to each row and each column.
When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa.
When a button attached to a column is pressed, the colors of all the squares in that column are inverted.

Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Constraints

1 \leq N,M \leq 1000

0 \leq K \leq NM

Input

Input is given from Standard Input in the following format:

N M K

Output

If Takahashi can have exactly K black squares in the grid, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

Press the buttons in the order of the first row, the first column.

Sample Input 2

2 2 1

Sample Output 2

No

Sample Input 3

3 5 8

Sample Output 3

Yes

Press the buttons in the order of the first column, third column, second row, fifth column.

Sample Input 4

7 9 20

Sample Output 4

No


Score : 100 points

Problem Statement

We have an N \times N square grid.

We will paint each square in the grid either black or white.

If we paint exactly A squares white, how many squares will be painted black?

Constraints

1 \leq N \leq 100

0 \leq A \leq N^2

Inputs

Input is given from Standard Input in the following format:

N
A

Outputs

Print the number of squares that will be painted black.

Sample Input 1

3
4

Sample Output 1

5

There are nine squares in a 3 \times 3 square grid.
Four of them will be painted white, so the remaining five squares will be painted black.

Sample Input 2

19
100

Sample Output 2

261

Sample Input 3

10
0

Sample Output 3

100

As zero squares will be painted white, all the squares will be painted black.


(let* ((n (read))
       (m (read))
       (lst (loop :repeat n :collect(read))))
  (format t "~A~%"
          (reduce #'+ (mapcar (lambda (k) (* 2 (min k (- m k)))) lst))))

Score : 200 points

Problem Statement

There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i).
Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N.

In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B.
Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i).
Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N.

When activated, each type of robot will operate as follows.

When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.

Constraints

1 \leq N \leq 100

1 \leq K \leq 100

0 < x_i < K

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
K
x_1 x_2 ... x_N

Outputs

Print the minimum possible total distance covered by robots.

Sample Input 1

1
10
2

Sample Output 1

4

There are just one ball, one type-A robot and one type-B robot.

If the type-A robot is used to collect the ball, the distance from the robot to the ball is 2, and the distance from the ball to the original position of the robot is also 2, for a total distance of 4.

Similarly, if the type-B robot is used, the total distance covered will be 16.

Thus, the total distance covered will be minimized when the type-A robot is used. The output should be 4.

Sample Input 2

2
9
3 6

Sample Output 2

12

The total distance covered will be minimized when the first ball is collected by the type-A robot, and the second ball by the type-B robot.

Sample Input 3

5
20
11 12 9 17 12

Sample Output 3

74


;; -*- 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
;;;
;;; Modular arithmetic
;;;

;; Blankinship algorithm
;; Reference: https://topcoder.g.hatena.ne.jp/spaghetti_source/20130126/1359171466 (Japanese)
(declaim (ftype (function * (values fixnum fixnum &optional)) %ext-gcd))
(defun %ext-gcd (a b)
  (declare (optimize (speed 3) (safety 0))
           (fixnum a b))
  (let ((y 1)
        (x 0)
        (u 1)
        (v 0))
    (declare (fixnum y x u v))
    (loop (when (zerop a)
            (return (values x y)))
          (let ((q (floor b a)))
            (decf x (the fixnum (* q u)))
            (rotatef x u)
            (decf y (the fixnum (* q v)))
            (rotatef y v)
            (decf b (the fixnum (* q a)))
            (rotatef b a)))))

;; recursive version
;; (defun %ext-gcd (a b)
;;   (declare (optimize (speed 3) (safety 0))
;;            (fixnum a b))
;;   (if (zerop b)
;;       (values 1 0)
;;       (multiple-value-bind (p q) (floor a b) ; a = pb + q
;;         (multiple-value-bind (v u) (%ext-gcd b q)
;;           (declare (fixnum u v))
;;           (values u (the fixnum (- v (the fixnum (* p u)))))))))

;; TODO: deal with bignums
(declaim (inline ext-gcd))
(defun ext-gcd (a b)
  "Returns two integers X and Y which satisfy AX + BY = gcd(A, B)."
  (declare ((integer #.(- most-positive-fixnum) #.most-positive-fixnum) a b))
  (if (>= a 0)
      (if (>= b 0)
          (%ext-gcd a b)
          (multiple-value-bind (x y) (%ext-gcd a (- b))
            (declare (fixnum x y))
            (values x (- y))))
      (if (>= b 0)
          (multiple-value-bind (x y) (%ext-gcd (- a) b)
            (declare (fixnum x y))
            (values (- x) y))
          (multiple-value-bind (x y) (%ext-gcd (- a) (- b))
            (declare (fixnum x y))
            (values (- x) (- y))))))

(declaim (inline mod-inverse)
         (ftype (function * (values (mod #.most-positive-fixnum) &optional)) mod-inverse))
;; (defun mod-inverse (a modulus)
;;   "Solves ax ≡ 1 mod m. A and M must be coprime."
;;   (declare (integer a)
;;            ((integer 1 #.most-positive-fixnum) modulus))
;;   (mod (%ext-gcd (mod a modulus) modulus) modulus))

(defun mod-inverse (a modulus)
  (declare ((integer 1 #.most-positive-fixnum) modulus))
  (let ((a (mod a modulus))
        (b modulus)
        (u 1)
        (v 0))
    (declare (fixnum a b u v))
    (loop until (zerop b)
          for quot = (floor a b)
          do (decf a (the fixnum (* quot b)))
             (rotatef a b)
             (decf u (the fixnum (* quot v)))
             (rotatef u v))
    (setq u (mod u modulus))
    (if (< u 0)
        (+ u modulus)
        u)))

;; not tested
;; TODO: move to another file
(declaim (inline binomial))
(defun mod-binomial (n k modulus)
  (declare ((integer 0 #.most-positive-fixnum) modulus))
  (if (or (< n k) (< n 0) (< k 0))
      0
      (let ((k (if (< k (- n k)) k (- n k)))
            (num 1)
            (denom 1))
        (declare ((integer 0) k num denom))
        (loop for x from n above (- n k)
              do (setq num (mod (* num x) modulus)))
        (loop for x from 1 to k
              do (setq denom (mod (* denom x) modulus)))
        (mod (* num (mod-inverse denom modulus)) modulus))))

(declaim (inline %calc-min-factor))
(defun %calc-min-factor (x alpha)
  "Returns k, so that x+k*alpha is the smallest non-negative number."
  (if (plusp alpha)
      (ceiling (- x) alpha)
      (floor (- x) alpha)))

(declaim (inline %calc-max-factor))
(defun %calc-max-factor (x alpha)
  "Returns k, so that x+k*alpha is the largest non-positive number."
  (if (plusp alpha)
      (floor (- x) alpha)
      (ceiling (- x) alpha)))

(defun solve-bezout (a b c &optional min max)
  "Returns an integer solution of a*x+b*y = c if it exists, otherwise
returns (VALUES NIL NIL).

If MIN is specified and MAX is null, the returned x is the smallest integer
equal or larger than MIN. If MAX is specified and MIN is null, x is the largest
integer equal or smaller than MAX. If the both are specified, x is an integer in
[MIN, MAX]. This function returns NIL when no x that satisfies the given
condition exists."
  (declare (fixnum a b c)
           ((or null fixnum) min max))
  (let ((gcd-ab (gcd a b)))
    (if (zerop (mod c gcd-ab))
        (multiple-value-bind (init-x init-y) (ext-gcd a b)
          (let* ((factor (floor c gcd-ab))
                 ;; m*x0 + n*y0 = d
                 (x0 (* init-x factor))
                 (y0 (* init-y factor)))
            (if (and (null min) (null max))
                (values x0 y0)
                (let (;; general solution: x = x0 + kΔx, y = y0 - kΔy
                      (deltax (floor b gcd-ab))
                      (deltay (floor a gcd-ab)))
                  (if min
                      (let* ((k-min (%calc-min-factor (- x0 min) deltax))
                             (x (+ x0 (* k-min deltax)))
                             (y (- y0 (* k-min deltay))))
                        (if (and max (> x max))
                            (values nil nil)
                            (values x y)))
                      (let* ((k-max (%calc-max-factor (- x0 max) deltax))
                             (x (+ x0 (* k-max deltax)))
                             (y (- y0 (* k-max deltay))))
                        (if (<= x max)
                            (values x y)
                            (values nil nil))))))))
        (values nil nil))))

(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* ((a (read))
         (b (read))
         (c (read))
         (d (read))
         (e (read))
         (f (read))
         (gcd-ab (gcd a b))
         (gcd-cd (gcd c d))
         (max 0)
         (arg-total 0)
         (arg-sugar 0)
         water-set
         sugar-set)
    (declare (uint16 a b c d e f))
    (loop for x from 1 to f
          do (multiple-value-bind (_ v) (solve-bezout a b x 0)
               (when (and v (>= v 0))
                 (push x water-set)))
             (multiple-value-bind (_ v) (solve-bezout c d x 0)
               (when (and v (>= v 0))
                 (push x sugar-set))))
    (dolist (water water-set)
      (dolist (sugar sugar-set)
        (declare (uint31 water sugar))
        (let* ((total (+ sugar (* 100 water)))
               (strength (/ sugar total)))
          (when (and (<= total f)
                     (<= sugar (* e water))
                     (> strength max))
            (setq max strength
                  arg-total total
                  arg-sugar sugar)))))
    (format t "~D ~D~%" arg-total arg-sugar)))

#-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 "1 2 10 20 15 200
"
    "110 10
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "1 2 1 2 100 1000
"
    "200 100
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "17 19 22 26 55 2802
"
    "2634 934
")))

Score : 300 points

Problem Statement

Snuke is making sugar water in a beaker.
Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations.

Operation 1: Pour 100A grams of water into the beaker.

Operation 2: Pour 100B grams of water into the beaker.

Operation 3: Put C grams of sugar into the beaker.

Operation 4: Put D grams of sugar into the beaker.

In our experimental environment, E grams of sugar can dissolve into 100 grams of water.

Snuke will make sugar water with the highest possible density.

The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker.
Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it.
If there is more than one candidate, any of them will be accepted.

We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent.
Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.

Constraints

1 \leq A < B \leq 30

1 \leq C < D \leq 30

1 \leq E \leq 100

100A \leq F \leq 3 000

A, B, C, D, E and F are all integers.

Inputs

Input is given from Standard Input in the following format:

A B C D E F

Outputs

Print two integers separated by a space.
The first integer should be the mass of the desired sugar water, and the second should be the mass of the sugar dissolved in it.

Sample Input 1

1 2 10 20 15 200

Sample Output 1

110 10

In this environment, 15 grams of sugar can dissolve into 100 grams of water, and the beaker can contain at most 200 grams of substances.

We can make 110 grams of sugar water by performing Operation 1 once and Operation 3 once.
It is not possible to make sugar water with higher density.
For example, the following sequences of operations are infeasible:

If we perform Operation 1 once and Operation 4 once, there will be undissolved sugar in the beaker.

If we perform Operation 2 once and Operation 3 three times, the mass of substances in the beaker will exceed 200 grams.

Sample Input 2

1 2 1 2 100 1000

Sample Output 2

200 100

There are other acceptable outputs, such as:

400 200

However, the output below is not acceptable:

300 150

This is because, in order to make 300 grams of sugar water containing 150 grams of sugar, we need to pour exactly 150 grams of water into the beaker, which is impossible.

Sample Input 3

17 19 22 26 55 2802

Sample Output 3

2634 934


(defvar +maxiter+ 50)

(let* ((a (read))
       (b (read))
       (c (read))
       (d (read))
       (e (read))
       (f (read))
       (rate 0)
       (max-sugar 0)
       (max-water 0))
  (dotimes (ai +maxiter+)
    (let ((wa (* a ai 100)))
      (when (< wa f)
        (dotimes (bi +maxiter+)
          (let ((wab (+ wa (* b bi 100))))
            (when (< 0 wab f)
              (dotimes (ci +maxiter+)
                (let ((sc (* c ci)))
                  (when (and (>= (* e wab) (* sc 100))
                             (>= f (+ sc wab)))
                    (dotimes (di +maxiter+)
                      (let ((scd (+ sc (* d di))))
                        (when (and (>= (* e wab) (* scd 100))
                                   (>= f (+ scd wab)))
                          (let ((current (/ scd (+ wab scd))))
                            (when (< rate current)
                              (setf rate current
                                    max-sugar scd
                                    max-water wab)))))))))))))))
  (format t "~A ~A~%"
          (+ max-sugar max-water)
          max-sugar))


(declaim (optimize (speed 3) (debug 0) (safety 0)))

;;cyclone-loop
(defun cyclone-reader (stream char)
  (declare (ignore char))
  `(loop ,@(read stream t nil t) do ,(read stream t nil t)))
(set-macro-character #\CYCLONE #'cyclone-reader)
;;

;;compiling-main
(defun main-reader (stream char)
  (declare (ignore char))
  `(funcall (compile nil (lambda () ,(read stream t nil t)))))
(set-macro-character #\HOUSE_BUILDING #'main-reader)
;;

🏠
(let ((a (* (read) 100))
      (b (* (read) 100))
      (c (read))
      (d (read))
      (e (read))
      (f (read)))
  (declare (fixnum a)
           (fixnum b)
           (fixnum c)
           (fixnum d)
           (fixnum e)
           (fixnum f))
  🌀 (for an to (truncate f a)
             :with max rational  = -1
             :with amax fixnum = -1
             :with smax fixnum = 0
             :finally (format t "~A ~A~%" amax smax))
  🌀 (for bn to (truncate (- f (* an a)) b))
  🌀 (for cn to (truncate (- f (+ (* an a) (* bn b))) c))
  🌀 (for dn to (truncate (- f (+ (* an a) (* bn b) (* cn c))) d)
             :for s fixnum = (+ (* cn c) (* dn d))
             :for w fixnum = (+ (* an a) (* bn b))
             :for all fixnum = (+ w s))
  (when (and (not (zerop all))
             (> (/ s all) max)
             (<= all f)
             (<= (* s 100) (* e w)))
    (setf max (/ s all))
    (setf amax all)
    (setf smax s)))


(princ(if(or(=(floor(setq a(read))10)9)(=(mod a 10)9))"Yes""No"))

Score : 100 points

Problem Statement

It is September 9 in Japan now.

You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

Constraints

10≤N≤99

Input

Input is given from Standard Input in the following format:

N

Output

If 9 is contained in the decimal notation of N, print Yes; if not, print No.

Sample Input 1

29

Sample Output 1

Yes

The one's digit of 29 is 9.

Sample Input 2

72

Sample Output 2

No

72 does not contain 9.

Sample Input 3

91

Sample Output 3

Yes


(print (if (find #\9 (read-line)) "Yes" "No"))


(let* ((n (read))
    (a (make-array n :initial-element 0))
    (beforenum 0)
    (times 0)
    (ans 0))

    (dotimes (i n)
        (setf (aref a i) (read))
    )
    (sort a #'<)

    (setq beforenum (aref a 0))
    (setq times 1)
    (loop for i from 1 to (- n 1) do
        (if (= (aref a i) beforenum)
            (incf times)
            (progn
                (if (not (zerop (rem times 2)))
                    (incf ans)
                )
                (setq beforenum (aref a i))
                (setq times 1)
            )
        )
    )
    (if (not (zerop (rem times 2)))
        (incf ans)
    )
    (princ ans)
)

Score : 300 points

Problem Statement

You are playing the following game with Joisino.

Initially, you have a blank sheet of paper.

Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times.

Then, you are asked a question: How many numbers are written on the sheet now?

The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

Constraints

1≤N≤100000

1≤A_i≤1000000000(=10^9)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print how many numbers will be written on the sheet at the end of the game.

Sample Input 1

3
6
2
6

Sample Output 1

1

The game proceeds as follows:

6 is not written on the sheet, so write 6.

2 is not written on the sheet, so write 2.

6 is written on the sheet, so erase 6.

Thus, the sheet contains only 2 in the end. The answer is 1.

Sample Input 2

4
2
5
5
2

Sample Output 2

0

It is possible that no number is written on the sheet in the end.

Sample Input 3

6
12
22
16
22
18
12

Sample Output 3

2


(defun input (n)
  (let ((a (make-array n)))
    (loop for i below n do (setf (aref a i) (1- (read))))
    a))

(defmacro w-f (a n)
  `(dotimes (k ,n) (dotimes (i ,n) (dotimes (j ,n)
    (if (> (aref ,a i j) (+ (aref ,a i k) (aref ,a k j)))
      (setf (aref ,a i j) (+ (aref ,a i k) (aref ,a k j))))))))

(defun solve (n r rn dn)
  (let ((ans 1000000000000)
        (used (make-array r :initial-element nil)))
    (labels
      ((dfs (x k dist)
         (if (= k r)
           (when (< dist ans) (setf ans dist))
           (progn (setf (aref used x) t)
                  (loop for i below r
                        unless (aref used i)
                        do (dfs i (1+ k) (+ dist (aref dn (aref rn x) (aref rn i)))))
                  (setf (aref used x) nil)))))
      (dotimes (i r)
        (dfs i 1 0)))
    ans))

(let* ((n (read))
       (m (read))
       (r (read))
       (rn (input r))
       (d (make-array `(,n ,n) :initial-element 10000000000)))
  (loop repeat m
        for a = (1- (read)) for b = (1- (read)) for c = (read)
        do (setf (aref d a b) c)
        do (setf (aref d b a) c))
  (w-f d n)
  (princ (solve n r rn d)))

Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town A_i and B_i and has a length of C_i.

Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N-1)/2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

r_i≠r_j (i≠j)

1≤A_i,B_i≤N, A_i≠B_i

(A_i,B_i)≠(A_j,B_j),(A_i,B_i)≠(B_j,A_j) (i≠j)

1≤C_i≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M R
r_1 ... r_R
A_1 B_1 C_1
:
A_M B_M C_M

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

3 3 3
1 2 3
1 2 1
2 3 1
3 1 4

Sample Output 1

2

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

3 3 2
1 3
2 3 2
1 3 6
1 2 2

Sample Output 2

4

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

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

Sample Output 3

3


Score : 100 points

Problem Statement

We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.

How many grams of sand will the upper bulb contains after t seconds?

Constraints

1≤X≤10^9

1≤t≤10^9

X and t are integers.

Input

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

X t

Output

Print the number of sand in the upper bulb after t second.

Sample Input 1

100 17

Sample Output 1

83

17 out of the initial 100 grams of sand will be consumed, resulting in 83 grams.

Sample Input 2

48 58

Sample Output 2

0

All 48 grams of sand will be gone, resulting in 0 grams.

Sample Input 3

1000000000 1000000000

Sample Output 3

0


(setq a (read))
(setq b (read))
(print (max 0 (- a b)))


(let* ((x! (read))
       (t! (read))
       (diff (- x! t!)))
    (if (> diff 0)
        (princ diff)
        (princ 0))
    (princ #\LineFeed))


(let ((s (concatenate 'list (read-line))))

  (defun f (lst)
    (if (null lst)
      nil
      (cons (car lst) (f (cddr lst)))))

  (format t "~A~%"
          (concatenate 'string (f s))))

Score : 200 points

Problem Statement

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

Constraints

Each character in s is a lowercase English letter.

1≤|s|≤10^5

Input

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

s

Output

Print the string obtained by concatenating all the characters in the odd-numbered positions.

Sample Input 1

atcoder

Sample Output 1

acdr

Extract the first character a, the third character c, the fifth character d and the seventh character r to obtain acdr.

Sample Input 2

aaaa

Sample Output 2

aa

Sample Input 3

z

Sample Output 3

z

Sample Input 4

fukuokayamaguchi

Sample Output 4

fkoaaauh


#-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))))
(defmacro dp (name args memo-size &body body)
  (let ((memo (gensym)))
    `(let ((,memo (make-array ,memo-size :initial-element nil)))
       (defun ,name ,args
         (or (aref ,memo ,@args)
             (setf (aref ,memo ,@args)
                   ,@body))))))

(defun split (x str)
  (let ((pos (search x str))
        (size (length x)))
    (if pos
        (cons (subseq str 0 pos)
              (split x (subseq str (+ pos size))))
        (list str))))

(defmacro collect-times (time body)
  `(loop repeat ,time collect ,body))

(defun read-times (time)
  (collect-times time (read)))

(defmacro aif (expr then else)
  `(let ((it ,expr))
     (if it ,then ,else)))

(defmacro awhen (expr &rest then)
  `(aif ,expr (progn ,@then) nil))

(defun comulative (function list &key base)
  (if base
      (comulative function (cons base list))
      (do ((lst (cdr list) (cdr lst))
           (acc (list (car list))))
          ((null lst) (reverse acc))
        (push (funcall function (car acc) (car lst)) acc))))

(defun binary-search (function left right
                      &optional
                        (eps 1)
                        (average-function (lambda (x y) (ash (+ x y) -1))))
  (if (<= (abs (- left right)) eps)
      right
      (let ((mid (funcall average-function left right)))
        (if (funcall function mid)
            (binary-search function left mid eps average-function)
            (binary-search function mid right eps average-function)))))

(defun arithmetic-mean (&rest body)
  (/ (apply #'+ body) (length body)))

(defvar +MOD+ (+ (expt 10 9) 7))

(defun mapa-b (fn a b &optional (step 1))
  (do ((i a (+ i step))
       (result nil))
      ((> i b) (nreverse result))
    (push (funcall fn i) result)))

(defun map0-n (fn n)
  (mapa-b fn 0 n))

(defun map1-n (fn n)
  (mapa-b fn 1 n))

(defun read-string (&optional (stream *standard-input*))
  (labels ((is-empty (x)
             (or (char= x #\space) (char= x #\newline))))
    (loop for char = (read-char stream)
          with result
          when (is-empty char)
            do (return (concatenate 'string (nreverse result)))
          do (push char result))))

(defun merge-sort (lst &optional (compare #'<))
  (let ((turn 0))
    (labels ((merge-list (a b a-length b-length)
               (cond ((zerop a-length) b)
                     ((zerop b-length) a)
                     ((funcall compare (car b) (car a))
                      (incf turn a-length)
                      (cons (car b)
                            (merge-list a (cdr b) a-length (1- b-length))))
                     (t
                      (cons (car a)
                            (merge-list (cdr a) b (1- a-length) b-length)))))
             (f (lst length)
               (if (= length 1)
                   lst
                   (let ((mid (ash length -1)))
                     (merge-list (f (subseq lst 0 mid) mid)
                                 (f (subseq lst mid) (- length mid))
                                 mid
                                 (- length mid))))))
      (values (f lst (length lst)) turn))))

(defun group (lst &optional (test #'eql) (key nil))
  (let ((table (make-hash-table :test test)))
    (mapc (lambda (x)
            (push x (gethash (if key (funcall key x) x) table)))
          lst)
    (loop for value being each hash-value in table
          collect value)))

(defun main (lst)
  (let ((group (sort (group (append (list -1000000000 9000000000) lst))
                     #'<
                     :key #'car)))
    (loop for i in group
          for j in (cdr group)
          for k in (cddr group)
          maximize (+ (length j)
                      (if (= (- (car j) (car i)) 1) (length i) 0)
                      (if (= (- (car k) (car j)) 1) (length k) 0)))))

(princ (main (read-times (read))))

Score : 300 points

Problem Statement

You are given an integer sequence of length N, a_1,a_2,...,a_N.

For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing.

After these operations, you select an integer X and count the number of i such that a_i=X.

Maximize this count by making optimal choices.

Constraints

1≤N≤10^5

0≤a_i<10^5 (1≤i≤N)

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 maximum possible number of i such that a_i=X.

Sample Input 1

7
3 1 4 1 5 9 2

Sample Output 1

4

For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.

Sample Input 2

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

3

Sample Input 3

1
99999

Sample Output 3

1


(let* ((n (1- (read)))
       (arr (make-array (1+ n) :initial-element 0 :element-type 'fixnum))
       (mem (make-array 100000 :initial-element 0 :element-type 'fixnum)))
  (loop :for a :from 0 :upto n :do(setf (aref arr a) (read)))
  (defun f (k) (if (= 0 (aref mem k))
                   (let ((q (count-if (lambda (a) (< -2 (- k a) 2)) arr)))
                     (setf (aref mem k) q)
                     q)
                   (aref mem k)))
  (princ (loop :for s :across arr :maximize(f s))))


;; -*- 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
(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* ((q (read))
         (h (read))
         (s (read))
         (d (read))
         (n (read)))
    (setq s (min (* 4 q) (* 2 h) s))
    (println
     (min (* s n)
          (+ (* d (floor n 2))
             (if (evenp n) 0 s))))))

#-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 "20 30 70 90
3
"
    "150
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "10000 1000 100 10
1
"
    "100
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "10 100 1000 10000
1
"
    "40
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "12345678 87654321 12345678 87654321
123456789
"
    "1524157763907942
")))

Score : 300 points

Problem Statement

You've come to your favorite store Infinitesco to buy some ice tea.

The store sells ice tea in bottles of different volumes at different costs.
Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen.
The store has an infinite supply of bottles of each type.

You want to buy exactly N liters of ice tea. How many yen do you have to spend?

Constraints

1 \leq Q, H, S, D \leq 10^8

1 \leq N \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q H S D
N

Output

Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.

Sample Input 1

20 30 70 90
3

Sample Output 1

150

Buy one 2-liter bottle and two 0.5-liter bottles. You'll get 3 liters for 90 + 30 + 30 = 150 yen.

Sample Input 2

10000 1000 100 10
1

Sample Output 2

100

Even though a 2-liter bottle costs just 10 yen, you need only 1 liter.
Thus, you have to buy a 1-liter bottle for 100 yen.

Sample Input 3

10 100 1000 10000
1

Sample Output 3

40

Now it's better to buy four 0.25-liter bottles for 10 + 10 + 10 + 10 = 40 yen.

Sample Input 4

12345678 87654321 12345678 87654321
123456789

Sample Output 4

1524157763907942


Score : 100 points

Problem Statement

Snuke lives at position x on a number line.
On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.

Snuke decided to get food delivery from the closer of stores A and B.
Find out which store is closer to Snuke's residence.

Here, the distance between two points s and t on a number line is represented by |s-t|.

Constraints

1 \leq x \leq 1000

1 \leq a \leq 1000

1 \leq b \leq 1000

x, a and b are pairwise distinct.

The distances between Snuke's residence and stores A and B are different.

Input

Input is given from Standard Input in the following format:

x a b

Output

If store A is closer, print A; if store B is closer, print B.

Sample Input 1

5 2 7

Sample Output 1

B

The distances between Snuke's residence and stores A and B are 3 and 2, respectively.
Since store B is closer, print B.

Sample Input 2

1 999 1000

Sample Output 2

A


(defun f (lst acc)
  (if lst
      (let ((head (car lst))
            (rest (cdr lst)))
        (if (find head rest)
            (f (remove head rest :count 1) (cons head acc))
            (f rest acc)))
      acc))

(let* ((n (read))
       (a (sort (loop repeat n collect (read)) #'<))
       (b (f a nil)))
  (princ (if (< (length b) 2)
             0
             (* (first b) (second b))))
  (terpri))

Score : 300 points

Problem Statement

We have N sticks with negligible thickness.
The length of the i-th stick is A_i.

Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides.
Find the maximum possible area of the rectangle.

Constraints

4 \leq N \leq 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible area of the rectangle.
If no rectangle can be formed, print 0.

Sample Input 1

6
3 1 2 4 2 1

Sample Output 1

2

1 \times 2 rectangle can be formed.

Sample Input 2

4
1 2 3 4

Sample Output 2

0

No rectangle can be formed.

Sample Input 3

10
3 3 3 3 4 4 4 5 5 5

Sample Output 3

20


(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 1))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (ql:quickload '(:cl-debug-print :fiveam)))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

;; BEGIN_INSERTED_CONTENTS
(defun ensure-list (x) (if (listp x) x (list x)))
(defmacro nlet (name args &body body)
  (let ((args (mapcar #'ensure-list args)))
    `(labels ((,name ,(mapcar #'car args) ,@body))
       (,name ,@(mapcar #'cadr args)))))

(defun split-to-list (str size)
  (declare (optimize (speed 3) (safety 1))
           (fixnum size)
           (string str))
  (loop for idx below size
        for pos1 = 0 then (1+ pos2)
        for pos2 = (position #\space str :start pos1 :test #'char=)
        collect (parse-integer str :start pos1 :end pos2)))

(defmacro println (obj &optional (stream '*standard-output*))
  `(let ((*read-default-float-format* 'double-float))
     (prog1 (princ ,obj ,stream) (terpri ,stream))))

;; Hauptteil
(defun extract-available-bars (list)
  (nlet recurse ((list list) (hash (make-hash-table)) res)
    (cond ((null list) res)
          ((gethash (car list) hash)
           (setf (gethash (car list) hash) nil)
           (recurse (cdr list) hash (cons (car list) res)))
          (t (setf (gethash (car list) hash) t)
             (recurse (cdr list) hash res)))))

(deftype uint nil `(integer 0 ,(expt 10 9)))
(defun main ()
  (let* ((n (read))
         (bars (split-to-list (read-line) n))
         (available-bars (extract-available-bars bars)))
    (if (null (cdr available-bars))
        (println 0)
        (let* ((longest (reduce #'max available-bars))
               (second-longest (reduce #'max (delete longest available-bars :count 1))))
          (println (* longest second-longest))))))

#-swank(main)


#-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))))

(let ((a (read))
	  (b (read))
	  (c (read))
	  (d (read)))
  (princ (max 0
			  (- (min b d) (max a c)))))

Score : 200 points

Problem Statement

Alice and Bob are controlling a robot. They each have one switch that controls the robot.

Alice started holding down her button A second after the start-up of the robot, and released her button B second after the start-up.

Bob started holding down his button C second after the start-up, and released his button D second after the start-up.

For how many seconds both Alice and Bob were holding down their buttons?

Constraints

0≤A<B≤100

0≤C<D≤100

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the length of the duration (in seconds) in which both Alice and Bob were holding down their buttons.

Sample Input 1

0 75 25 100

Sample Output 1

50

Alice started holding down her button 0 second after the start-up of the robot, and released her button 75 second after the start-up.

Bob started holding down his button 25 second after the start-up, and released his button 100 second after the start-up.

Therefore, the time when both of them were holding down their buttons, is the 50 seconds from 25 seconds after the start-up to 75 seconds after the start-up.

Sample Input 2

0 33 66 99

Sample Output 2

0

Alice and Bob were not holding their buttons at the same time, so the answer is zero seconds.

Sample Input 3

10 90 20 80

Sample Output 3

60


(let((a(read)))(princ(max(-(min(read)(+(*(setq c(read))0)(read)))(max a c))0)))


(defun read-prob ()
  (let ((lst nil))
    (dotimes (cnt 4 (nreverse lst))
      (push (read) lst)
      )))

(defun solve (lst)
  (let* ((st (max (nth 0 lst) (nth 2 lst)))
         (en (min (nth 1 lst) (nth 3 lst)))
         (ans (- en st)))
    (if (< ans 0)
        0 ans)))

(format t "~A" (solve (read-prob)))


(let ((n (read))
       (result 1))
  (dotimes (i n)
    (setf result (lcm result (read))))
  (format t "~a~%" result))

Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.

Initially, the hand of every clock stands still, pointing directly upward.

Now, Dolphin starts all the clocks simultaneously.

In how many seconds will the hand of every clock point directly upward again?

Constraints

1≤N≤100

1≤T_i≤10^{18}

All input values are integers.

The correct answer is at most 10^{18} seconds.

Input

Input is given from Standard Input in the following format:

N
T_1
:
T_N

Output

Print the number of seconds after which the hand of every clock point directly upward again.

Sample Input 1

2
2
3

Sample Output 1

6

We have two clocks. The time when the hand of each clock points upward is as follows:

Clock 1: 2, 4, 6, ... seconds after the beginning

Clock 2: 3, 6, 9, ... seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

Sample Input 2

5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

1000000000000000000


Score : 100 points

Problem Statement

In K-city, there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?

Constraints

2 ≤ n, m ≤ 100

Input

Input is given from Standard Input in the following format:

n m

Output

Print the number of blocks in K-city.

Sample Input 1

3 4

Sample Output 1

6

There are six blocks, as shown below:

Sample Input 2

2 2

Sample Output 2

1

There are one block, as shown below:


(let* ((s (concatenate 'list (read-line)))
       (f (string (car s)))
       (b (string (car (reverse s)))))

  (format t "~A~%"
          (concatenate 'string f (write-to-string (- (length s) 2)) b)))

Score : 200 points

Problem Statement

The word internationalization is sometimes abbreviated to i18n.
This comes from the fact that there are 18 letters between the first i and the last n.

You are given a string s of length at least 3 consisting of lowercase English letters.
Abbreviate s in the same way.

Constraints

3 ≤ |s| ≤ 100 (|s| denotes the length of s.)

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the abbreviation of s.

Sample Input 1

internationalization

Sample Output 1

i18n

Sample Input 2

smiles

Sample Output 2

s4s

Sample Input 3

xyz

Sample Output 3

x1z


;; -*- 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
(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
                                  (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))
         (as (loop repeat n collect (read-fixnum)))
         (num2mod4 (count-if (lambda (a) (= 2 (mod a 4))) as)))
    (declare (uint32 n) (list as))
    (write-line
     (if (<= num2mod4 1)
         (if (>= (count-if (lambda (a) (zerop (mod a 4))) as)
                 (max 1 (floor (length as) 2)))
             "Yes"
             "No")
         (let ((as (delete-if (lambda (a) (= 2 (mod a 4))) as)))
           (if (null as)
               "Yes"
               (if (>= (count-if (lambda (a) (zerop (mod a 4))) as)
                       (max 1 (floor (length as) 2)))
                   "Yes"
                   "No")))))))

#-swank(main)

;; For Test
#+swank
(defun io-equal (in-string out-string &optional (func #'main))
  "Passes IN-STRING to *STANDARD-INPUT*, executes FUNC, and returns true if the
string output to *STANDARD-OUTPUT* is equal to OUT-STRING."
  (labels ((ensure-last-lf (s)
             (if (and (> (length s) 0)
                      (eql (char s (- (length s) 1)) #\Linefeed))
                 s
                 (uiop:strcat s uiop:+lf+))))
    (equal (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 func)))))))

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    (run-program "C:/msys64/usr/bin/cat.exe" '("/dev/clipboard") :output out)))

#+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*))
  (let ((*standard-output* out))
    (etypecase thing
      (null ; Runs #'MAIN with the string on clipboard
       (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 ; Runs #'MAIN with the string in a text file
       (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
1 10 100
"
    "Yes
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "4
1 2 3 4
"
    "No
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
1 4 1
"
    "Yes
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
1 1
"
    "No
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "6
2 7 1 8 2 8
"
    "Yes
")))

Score : 400 points

Problem Statement

We have a sequence of length N, a = (a_1, a_2, ..., a_N).
Each a_i is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

2 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.

Sample Input 1

3
1 10 100

Sample Output 1

Yes

One solution is (1, 100, 10).

Sample Input 2

4
1 2 3 4

Sample Output 2

No

It is impossible to permute a so that the condition is satisfied.

Sample Input 3

3
1 4 1

Sample Output 3

Yes

The condition is already satisfied initially.

Sample Input 4

2
1 1

Sample Output 4

No

Sample Input 5

6
2 7 1 8 2 8

Sample Output 5

Yes


(format t "~A~%" (expt 2 (floor (log (read) 2))))

Score : 200 points

Problem Statement

Takahashi loves numbers divisible by 2.

You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.

Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.

For example,

6 can be divided by 2 once: 6 -> 3.

8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.

3 can be divided by 2 zero times.

Constraints

1 ≤ N ≤ 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

7

Sample Output 1

4

4 can be divided by 2 twice, which is the most number of times among 1, 2, ..., 7.

Sample Input 2

32

Sample Output 2

32

Sample Input 3

1

Sample Output 3

1

Sample Input 4

100

Sample Output 4

64


(let* ((n (read))
       (m (read))
       (lst (loop :repeat m :collect (cons (read) (read)))))
  (if (loop :for k :from 2 :upto n :never (and (position (cons 1 k) lst) (position (cons k n) lst)))
      (princ "POSSIBLE")
      (princ "IMPOSSIBLE")))

Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

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

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

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

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.


(let* ((n (read))
       (m (read))
       (lst (loop :repeat m :collect (cons (read) (read)))))
  (if (loop :for k :from 2 :upto n :never (position (cons 1 k) lst) (position (cons k n) lst))
      (princ "POSSIBLE")
      (princ "IMPOSSIBLE")))


(declaim (optimize (speed 3) (debug 0) (safety 0)))

(defun find-path (start end path-list cnt)
  (if (and (< cnt 3) (= start end))
      cnt
      (let ((possible-path-list
              (remove-if-not (lambda (path)
                               (= (car path) start))
                             path-list)))
        (mapcar (lambda (path)
                  (find-path (cadr path) end path-list (1+ cnt)))
                possible-path-list))))

(defun reach? (tree)
  (let ((possible? nil))
    (labels ((walk-tree (tree)
               (cond (possible? nil)
                     ((and (numberp tree)
                           (<= tree 2))
                      (setf possible? t))
                     ((consp tree)
                      (walk-tree (car tree))
                      (walk-tree (cdr tree)))
                     ((null tree) nil))))
      (walk-tree tree)
      possible?)))

(let* ((n (read))
       (m (read))
       (path-list (loop repeat m collect (list (read) (read)))))
  (if (reach? (find-path 1 n path-list 0))
      (format t "POSSIBLE~%")
      (format t "IMPOSSIBLE~%")))


(defun find-path (start end path-list cnt)
  (if (and (< cnt 3) (= start end))
      cnt
      (let ((possible-path-list
              (remove-if-not (lambda (path)
                               (= (car path) start))
                             path-list)))
        (mapcar (lambda (path)
                  (find-path (cadr path) end path-list (1+ cnt)))
                possible-path-list))))

(defun reach? (tree)
  (let ((possible? nil))
    (labels ((walk-tree (tree)
               (cond (possible? nil)
                     ((and (numberp tree)
                           (<= tree 2))
                      (setf possible? t))
                     ((consp tree)
                      (walk-tree (car tree))
                      (walk-tree (cdr tree)))
                     ((null tree) nil))))
      (walk-tree tree)
      possible?)))

(let* ((n (read))
       (m (read))
       (path-list (loop repeat m collect (list (read) (read)))))
  (if (reach? (find-path 1 n path-list 0))
      (format t "POSSIBLE~%")
      (format t "IMPOSSIBLE~%")))


;; -*- 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
(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 feasible-p (as threshold)
  (declare ((simple-array uint16 (* *)) as)
           (uint32 threshold))
  #>threshold
  (destructuring-bind (n m) (array-dimensions as)
    (let ((indices (make-array n :element-type 'uint16 :initial-element 0))
          (marked (make-array m :element-type 'bit :initial-element 0))
          (table (make-array m :element-type 'uint16)))
      (labels ((step-index (i)
                 (loop (incf (aref indices i))
                       (when (>= (aref indices i) m)
                         (return-from feasible-p nil))
                       (when (zerop (aref marked (aref as i (aref indices i))))
                         (return)))))
        (loop
          (fill table 0)
          (dotimes (i n)
            (incf (aref table (aref as i (aref indices i)))))
          (let ((feasible t))
            (dotimes (i n)
              (when (> (aref table (aref as i (aref indices i))) threshold)
                (setf feasible nil)
                (setf (aref marked (aref as i (aref indices i))) 1)
                (step-index i)
                (when (>= (aref indices i) m)
                  (return-from feasible-p nil))))
            (when feasible
              (return-from feasible-p t))))))))

(defun main ()
  (let* ((n (read))
         (m (read))
         (as (make-array (list n m) :element-type 'uint16)))
    (declare (uint16 n m))
    (dotimes (i n)
      (dotimes (j m)
        (setf (aref as i j) (- (read-fixnum) 1))))
    (sb-int:named-let bisect ((ng 0) (ok m))
      (if (<= (- ok ng) 1)
          (println ok)
          (let ((mid (ash (+ ng ok) -1)))
            (if (feasible-p as mid)
                (bisect ng mid)
                (bisect mid ok)))))))

#-swank (main)

Score : 700 points

Problem Statement

Takahashi is hosting an sports meet.
There are N people who will participate. These people are conveniently numbered 1 through N.
Also, there are M options of sports for this event. These sports are numbered 1 through M.
Among these options, Takahashi will select one or more sports (possibly all) to be played in the event.

Takahashi knows that Person i's j-th favorite sport is Sport A_{ij}.
Each person will only participate in his/her most favorite sport among the ones that are actually played in the event, and will not participate in the other sports.

Takahashi is worried that one of the sports will attract too many people.
Therefore, he would like to carefully select sports to be played so that the number of the participants in the sport with the largest number of participants is minimized.
Find the minimum possible number of the participants in the sport with the largest number of participants.

Constraints

1 \leq N \leq 300

1 \leq M \leq 300

A_{i1} , A_{i2} , ... , A_{iM} is a permutation of the integers from 1 to M.

Input

Input is given from Standard Input in the following format:

N M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
:
A_{N1} A_{N2} ... A_{NM}

Output

Print the minimum possible number of the participants in the sport with the largest number of participants.

Sample Input 1

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

Sample Output 1

2

Assume that Sports 1, 3 and 4 are selected to be played. In this case, Person 1 will participate in Sport 1, Person 2 in Sport 3, Person 3 in Sport 3 and Person 4 in Sport 4.
Here, the sport with the largest number of participants is Sport 3, with two participants.
There is no way to reduce the number of participants in the sport with the largest number of participants to 1. Therefore, the answer is 2.

Sample Input 2

3 3
2 1 3
2 1 3
2 1 3

Sample Output 2

3

Since all the people have the same taste in sports, there will be a sport with three participants, no matter what sports are selected.
Therefore, the answer is 3.


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

  (format t "~A~%"
          (if (or (zerop (mod a 3)) (zerop (mod b 3)) (zerop (mod (+ a b) 3)))
            "Possible"
            "Impossible")))

Score : 100 points

Problem Statement

Snuke is giving cookies to his three goats.

He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).

Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Constraints

1 \leq A,B \leq 100

Both A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If it is possible to give cookies so that each of the three goats can have the same number of cookies, print Possible; otherwise, print Impossible.

Sample Input 1

4 5

Sample Output 1

Possible

If Snuke gives nine cookies, each of the three goats can have three cookies.

Sample Input 2

1 1

Sample Output 2

Impossible

Since there are only two cookies, the three goats cannot have the same number of cookies no matter what Snuke gives to them.


(let* ((n (read))
       (k (read))
       (lst (sort (loop repeat n
                        collect (read))
                  #'>)))

  (format t "~A~%"
          (reduce #'+ (subseq lst 0 k))))

Score : 200 points

Problem Statement

Snuke has N sticks.
The length of the i-th stick is l_i.

Snuke is making a snake toy by joining K of the sticks together.

The length of the toy is represented by the sum of the individual sticks that compose it.
Find the maximum possible length of the toy.

Constraints

1 \leq K \leq N \leq 50

1 \leq l_i \leq 50

l_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
l_1 l_2 l_3 ... l_{N}

Output

Print the answer.

Sample Input 1

5 3
1 2 3 4 5

Sample Output 1

12

You can make a toy of length 12 by joining the sticks of lengths 3, 4 and 5, which is the maximum possible length.

Sample Input 2

15 14
50 26 27 21 41 7 42 35 7 5 5 36 39 1 45

Sample Output 2

386


(let* ((n (read))
       (lst (loop :repeat n :collect (read)))
       (mem (cons (reduce #'+ (cdr lst)) (car lst))))
  (princ (loop :for k :in (subseq lst 1 (1- n)) :minimize (abs (- (setf (car mem) (- (car mem) k))
                                                                  (setf (cdr mem) (+ (cdr mem) k)))))))

Score : 300 points

Problem Statement

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.

They will share these cards.
First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards.
Here, both Snuke and Raccoon have to take at least one card.

Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively.
They would like to minimize |x-y|.
Find the minimum possible value of |x-y|.

Constraints

2 \leq N \leq 2 \times 10^5

-10^{9} \leq a_i \leq 10^{9}

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print the answer.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

1

If Snuke takes four cards from the top, and Raccoon takes the remaining two cards, x=10, y=11, and thus |x-y|=1. This is the minimum possible value.

Sample Input 2

2
10 -10

Sample Output 2

20

Snuke can only take one card from the top, and Raccoon can only take the remaining one card. In this case, x=10, y=-10, and thus |x-y|=20.


;; -*- 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

(defun main ()
  (declare #.OPT)
  (let* ((n (read))
         (as (make-array n :element-type 'int32)))
    (dotimes (i n) (setf (aref as i) (read-fixnum)))
    (let* ((snuke (reduce #'+ as :start 1))
           (racoon (aref as 0))
           (res (abs (- snuke racoon))))
      (declare (fixnum snuke racoon res))
      (loop for i from 1 below (- n 1)
            do (decf snuke (aref as i))
               (incf racoon (aref as i))
               (setf res (min res (abs (- snuke racoon)))))
      (println res))))

#-swank(main)