(defun count2div (n)
    (labels ((f (n i)
                (if (not (zerop (rem n 2))) i
                    (f (floor n 2) (1+ i)))))
        (f n 0)))

(defun main ()
    (let* ((n (read))
           (a-list (loop for i below n collect (read))))
        (format t "~A~%"
            (reduce 'min (map 'list #'count2div a-list)))))

(main)

Score : 200 points

Problem Statement

There are N positive integers written on a blackboard: A_1, ..., A_N.

Snuke can perform the following operation when all integers on the blackboard are even:

Replace each integer X on the blackboard by X divided by 2.

Find the maximum possible number of operations that Snuke can perform.

Constraints

1 \leq N \leq 200

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of operations that Snuke can perform.

Sample Input 1

3
8 12 40

Sample Output 1

2

Initially, [8, 12, 40] are written on the blackboard.
Since all those integers are even, Snuke can perform the operation.

After the operation is performed once, [4, 6, 20] are written on the blackboard.
Since all those integers are again even, he can perform the operation.

After the operation is performed twice, [2, 3, 10] are written on the blackboard.
Now, there is an odd number 3 on the blackboard, so he cannot perform the operation any more.

Thus, Snuke can perform the operation at most twice.

Sample Input 2

4
5 6 8 10

Sample Output 2

0

Since there is an odd number 5 on the blackboard already in the beginning, Snuke cannot perform the operation at all.

Sample Input 3

6
382253568 723152896 37802240 379425024 404894720 471526144

Sample Output 3

8


(defun shift-only (n a)
  (let ((res 0)
        (exist-odd nil))
    (loop
       (dotimes (i n)
         (if (not (zerop (rem (aref a i) 2)))
             (setf exist-odd t)))
       (if (eq exist-odd t) (return))
       (dotimes (i n)
         (setf (aref a i) (/ (aref a i) 2)))
       (incf res))
    res))

(defun create-data ()
  (let* ((n (read))
         (a (make-array `(,n))))
    (dotimes (i n)
      (setf (aref a i) (read)))
    (values n a)))

(multiple-value-bind (n a) (create-data)
  (format t "~A~%" (shift-only n a)))

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

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.


(defun group (lst)
  (labels ((inner (lst k rev)
             (cond
               ((null lst) rev)
               ((not (equal (car lst) (cadr lst)))
                (inner (cdr lst) 1 (cons `(,(car lst) ,k) rev)))
               (t
                (inner (cdr lst) (1+ k) rev)))))
    (inner lst 1 '())))

(defun read-input (N)
  (sort (loop for n from 1 upto N collect (read))
        #'<))

(defun answer (lst k)
  (reduce #'(lambda (n p) (+ n (cadr p)))
          (subseq (sort
                   (group (sort lst #'<))
                   #'> :key #'cadr)
                  k)
          :initial-value 0))

(defun main ()
  (let ((N (read)) (K (read)))))
    (format t "~a~%"
            ))))
(main)

Score : 300 points

Problem Statement

Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.

He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.

Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Constraints

1 \leq K \leq N \leq 200000

1 \leq A_i \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

Print the minimum number of balls that Takahashi needs to rewrite the integers on them.

Sample Input 1

5 2
1 1 2 2 5

Sample Output 1

1

For example, if we rewrite the integer on the fifth ball to 2, there are two different integers written on the balls: 1 and 2.
On the other hand, it is not possible to rewrite the integers on zero balls so that there are at most two different integers written on the balls, so we should print 1.

Sample Input 2

4 4
1 1 2 2

Sample Output 2

0

Already in the beginning, there are two different integers written on the balls, so we do not need to rewrite anything.

Sample Input 3

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

Sample Output 3

3


;;; Utils (quoted from https://competitive12.blogspot.com/2020/03/common-lisp.html)

;; Read fixnum

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare (inline read-byte)
           #-swank (sb-kernel:ansi-stream in))
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (read-byte in nil 0))))
    (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) (the (integer 0 #.(floor most-positive-fixnum 10)) (* result 10))))
              (return (if minus (- result) result))))))))

;; fast read-line

(defmacro buffered-read-line (&optional (buffer-size 30) (in '*standard-input*) (term-char #\Space))
  (let ((buffer (gensym))
        (character (gensym))
        (idx (gensym)))
    `(let* ((,buffer (load-time-value (make-string ,buffer-size :element-type 'base-char))))
       (declare (simple-base-string ,buffer)
                (inline read-byte))
       (loop for ,character of-type base-char =
             ,(if (member :swank *features*)
                  `(read-char ,in nil #\Newline) ; on SLIME
                `(code-char (read-byte ,in nil #.(char-code #\Newline))))
             for ,idx from 0
             until (char= ,character #\Newline)
             do (setf (schar ,buffer ,idx) ,character)
             finally (when (< ,idx ,buffer-size)
                       (setf (schar ,buffer ,idx) ,term-char))
             (return (values ,buffer ,idx))))))

;;; invoke child process

#-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))))

;-------------------
; Write code here
;-------------------

(defun solve (n k a)
  (labels ((counter (a &optional (res (make-array (1+ n))))
             (if (null a)
                 (sort (remove 0 (coerce res 'list)) #'<=)
                 (progn
                   (incf (aref res (first a)))
                   (counter (rest a) res)))))
    (let* ((a-lst (counter a))
           (variation (length a-lst)))
      (labels ((counter-2 (variation k a-lst &optional (res 0))
                 (if (<= variation k)
                     res
                     (counter-2 (1- variation)
                                k
                                (rest a-lst)
                                (+ res (first a-lst))))))
        (counter-2 variation k a-lst)))))

(defun main ()
  (let* ((n (read))
         (k (read))
         (a (read-from-string (concatenate 'string "(" (read-line) ")"))))
    (princ (solve n k (sort a #'<=)))
    (fresh-line)))

(main)


(defun f (t a b) (min (* t a) b))
(print (f (read) (read) (read)))

Score : 100 points

Problem Statement

You are parking at a parking lot. You can choose from the following two fee plans:

Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours.

Plan 2: The fee will be B yen, regardless of the duration.

Find the minimum fee when you park for N hours.

Constraints

1≤N≤20

1≤A≤100

1≤B≤2000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

When the minimum fee is x yen, print the value of x.

Sample Input 1

7 17 120

Sample Output 1

119

If you choose Plan 1, the fee will be 7×17=119 yen.

If you choose Plan 2, the fee will be 120 yen.

Thus, the minimum fee is 119 yen.

Sample Input 2

5 20 100

Sample Output 2

100

The fee might be the same in the two plans.

Sample Input 3

6 18 100

Sample Output 3

100


(let* ((n (read))
    (a 0)
    (b n))

    (loop for i from 1 while (> b 0) do
        (progn
            (incf a (rem b 10))
            (setq b (floor (/ b 10)))
        )
    )
    (if (zerop (rem n a))
        (princ "Yes")
        (princ "No")
    )
)

Score : 200 points

Problem Statement

An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.

Given an integer N, determine whether it is a Harshad number.

Constraints

1?N?10^8

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print Yes if N is a Harshad number; print No otherwise.

Sample Input 1

12

Sample Output 1

Yes

f(12)=1+2=3. Since 12 is divisible by 3, 12 is a Harshad number.

Sample Input 2

57

Sample Output 2

No

f(57)=5+7=12. Since 57 is not divisible by 12, 12 is not a Harshad number.

Sample Input 3

148

Sample Output 3

No


(let* ((n (read-line))
       (fx (loop for ch across n sum (- (char-code ch) (char-code #\0)))))
  (if (= 0 (mod (parse-integer n) fx))
      (format t "Yes~%")
      (format t "NO~%")))


#|
------------------------------------
|               Utils               |
------------------------------------
|#

(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)

#-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))))

(defconstant +mod+ 1000000007)

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~a" b)) () '(signed-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~a" b)) () '(unsigned-byte ,b))) bits)))

(define-int-types 2 4 8 16 32 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1
        (write obj :stream stream)
      (fresh-line stream))))

(defmacro safe-sort (list test &key (key #'identity))
  `(progn
    (declaim (inline sort sb-impl::stable-sort-list))
    (sort (copy-seq ,list) ,test :key ,key)))

(defmacro read-numbers-to-list (size)
  `(loop repeat ,size collect (read)))

(defmacro read-numbers-to-array (size)
  (let ((i (gensym))
        (arr (gensym)))
    `(let ((,arr (make-array ,size
                             :element-type 'fixnum)))
       (declare ((array fixnum 1) ,arr))
       (loop for ,i of-type fixnum below ,size do
            (setf (aref ,arr ,i) (read))
          finally
            (return ,arr)))))

(defmacro read-characters-to-board (row-size column-size)
  (let ((board (gensym))
        (r (gensym))
        (c (gensym))
        (tmp (gensym)))
    `(let ((,board (make-array '(,row-size ,column-size) :element-type 'character :adjustable nil)))
       (dotimes (,r ,row-size ,board)
         (let ((,tmp (read-line)))
           (dotimes (,c ,column-size)
             (setf (aref ,board ,r ,c) (char ,tmp ,c))))))))

(defmethod princ-for-each-line ((sequence list))
  (format t "~{~a~&~}" sequence))

(defmethod princ-for-each-line ((sequence vector))
  (loop for i below (length sequence) do
       (princ (aref sequence i))
       (fresh-line)))

(declaim (inline unwrap))
(defun unwrap (list)
  (the string
       (format nil "~{~a~^ ~}" list)))

(defmacro with-buffered-stdout (&body body)
  (let ((out (gensym)))
    `(let ((,out (make-string-output-stream :element-type 'base-char)))
       (let ((*standard-output* ,out))
         ,@body)
       (write-string (get-output-stream-string ,out)))))

(defmacro maxf (place cand)
  `(setf ,place (max ,place ,cand)))

(defmacro minf (place cand)
  `(setf ,place (min ,place ,cand)))

#|
------------------------------------
|               Body               |
------------------------------------
|#

(defun calc-benefit (n choice f p)
  (loop for store below n
     with res = 0
     do
       (loop for time below 10
          with s = 0
          do
            (incf s (logand (ash choice (- time))
                            (aref f store time)))
          finally
            (incf res (aref p store s)))
     finally
       (return res)))

(defun solve (n f p)
  (loop for i from 1 below (ash 1 10)
       maximize (calc-benefit n i f p)))

(defun main ()
  (declare #.OPT)
  (let ((n (read)))
    (let ((f (make-array (list n 10)
                         :element-type 'fixnum))
          (p (make-array (list n 11)
                         :element-type 'fixnum)))
      (dotimes (i n)
        (dotimes (j 10)
          (setf (aref f i j) (read))))
      (dotimes (i n)
        (dotimes (j 11)
          (setf (aref p i j) (read))))
      (println (solve n f p)))))

#-swank (main)

Score : 300 points

Problem Statement

Joisino is planning to open a shop in a shopping street.

Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.

There are already N stores in the street, numbered 1 through N.

You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.

Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.

Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

Constraints

1≤N≤100

0≤F_{i,j,k}≤1

For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.

-10^7≤P_{i,j}≤10^7

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}

Output

Print the maximum possible profit of Joisino's shop.

Sample Input 1

1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2

Sample Output 1

8

If her shop is open only during the periods when Shop 1 is opened, the profit will be 8, which is the maximum possible profit.

Sample Input 2

2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1

Sample Output 2

-2

Note that a shop must be open during at least one period, and the profit may be negative.

Sample Input 3

3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7

Sample Output 3

23


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

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

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

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

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

(defconstant +mod+ 1000000007)

;; Body

(defun main ()
  (let* ((n (read))
         (c (read))
         (timelines (make-array (list 30 100002) :element-type 'bit :initial-element 0)))
    (declare (uint32 n) (ignore c))
    (dotimes (i n)
      (let ((begin (read-fixnum))
            (end (read-fixnum))
            (channel (- (read-fixnum) 1)))
        (loop for time from begin to end
              do (setf (aref timelines channel time) 1))))
    (println (loop for time below 100002
                   maximize (loop for channel below 30
                                  sum (aref timelines channel time) of-type uint32)))))

#-swank(main)

Score : 400 points

Problem Statement

Joisino is planning to record N TV programs with recorders.

The TV can receive C channels numbered 1 through C.

The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i.

Here, there will never be more than one program that are broadcast on the same channel at the same time.

When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T).

Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.

Constraints

1≤N≤10^5

1≤C≤30

1≤s_i<t_i≤10^5

1≤c_i≤C

If c_i=c_j and i≠j, either t_i≤s_j or s_i≥t_j.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N C
s_1 t_1 c_1
:
s_N t_N c_N

Output

When the minimum required number of recorders is x, print the value of x.

Sample Input 1

3 2
1 7 2
7 8 1
8 12 1

Sample Output 1

2

Two recorders can record all the programs, for example, as follows:

With the first recorder, record Channel 2 from time 1 to time 7. The first program will be recorded. Note that this recorder will be unable to record other channels from time 0.5 to time 7.

With the second recorder, record Channel 1 from time 7 to time 12. The second and third programs will be recorded. Note that this recorder will be unable to record other channels from time 6.5 to time 12.

Sample Input 2

3 4
1 3 2
3 4 4
1 4 3

Sample Output 2

3

There may be a channel where there is no program to record.

Sample Input 3

9 4
56 60 4
33 37 2
89 90 3
32 43 1
67 68 3
49 51 3
31 32 3
70 71 1
11 12 3

Sample Output 3

2


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

;; BEGIN_INSERTED_CONTENTS
(defun map-run-length-rotative (function seq &key (test #'eql))
  (declare (sequence seq)
           (function test function))
  (etypecase seq
    (vector
     (unless (zerop (length seq))
       (let ((prev (aref seq 0))
             (start 0)
             (init-elm (aref seq 0))
             (init-length 0))
         (declare ((integer 0 #.most-positive-fixnum) start init-length))
         (loop for pos from 1 below (length seq)
               unless (funcall test prev (aref seq pos))
               do (if (zerop init-length)
                      (setf init-length (- pos start))
                      (funcall function prev (- pos start)))
                  (setf prev (aref seq pos)
                        start pos)
               finally (if (funcall test init-elm (aref seq (- pos 1)))
                           (funcall function init-elm (+ init-length (- pos start)))
                           (progn
                             (funcall function prev (- pos start))
                             (funcall function init-elm init-length)))))))
    (list
     (error "Not implemented."))))

(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 calc (table)
  ;; (format t "~A~%" table)
  (let ((res #xffff))
    (declare (uint32 res))
    (map-run-length-rotative
     (lambda (x num)
       (declare (uint32 x num))
       (if (zerop x)
           (setq res (min res (+ num 1)))
           (when (> num 1)
             (setq res 1))))
     table)
    res))

(defun main ()
  (let* ((n (read))
         (counts (make-array 13 :element-type 'uint8 :initial-element 0))
         (res 0))
    (declare (uint8 res))
    (dotimes (i n) (incf (aref counts (read))))
    (incf (aref counts 0)) ; Takahashi-kun
    (when (>= (aref counts 0) 2)
      (println 0)
      (return-from main))
    (when (>= (aref counts 12) 2)
      (println 0)
      (return-from main))
    (loop for d from 1 to 11
          when (> (aref counts d) 2)
          do (println 0)
             (return-from main))
    (dotimes (bits (ash 1 11))
      (let ((table (make-array 24 :element-type 'bit :initial-element 0)))
        (declare (dynamic-extent table))
        (when (= (aref counts 0) 1)
          (setf (aref table 0) 1))
        (when (= (aref counts 12) 1)
          (setf (aref table 12) 1))
        (loop for d from 1 to 11
              when (= (aref counts d) 2)
              do (setf (aref table d) 1
                       (aref table (- 24 d)) 1))
        (loop for d from 1 to 11
              when (= (aref counts d) 1)
              do (if (logbitp (- d 1) bits)
                     (setf (aref table d) 1)
                     (setf (aref table (- 24 d)) 1)))
        (setf res (max res (calc table)))))
    (println res)))

#-swank (main)

Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.


(let ((l (concatenate 'list (read-line))))
  (format t "~A~%"
          (if (<= (length (remove-duplicates l)) 2)
            "Yes"
            "No")))

Score : 100 points

Problem Statement

We call a 4-digit integer with three or more consecutive same digits, such as 1118, good.

You are given a 4-digit integer N. Answer the question: Is N good?

Constraints

1000 ≤ N ≤ 9999

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is good, print Yes; otherwise, print No.

Sample Input 1

1118

Sample Output 1

Yes

N is good, since it contains three consecutive 1.

Sample Input 2

7777

Sample Output 2

Yes

An integer is also good when all the digits are the same.

Sample Input 3

1234

Sample Output 3

No


(let ((n (read)))

  (defun lucas (x &optional (t0 2) (t1 1))
    (if (= x 0)
      t0
      (if (= x 1)
        t1
        (+ (lucas (- x 2)) (lucas (1- x))))))

  (format t "~A~%"
          (lucas n)))

Score : 200 points

Problem Statement

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.

You are given an integer N. Find the N-th Lucas number.

Here, the i-th Lucas number L_i is defined as follows:

L_0=2

L_1=1

L_i=L_{i-1}+L_{i-2} (i≥2)

Constraints

1≤N≤86

It is guaranteed that the answer is less than 10^{18}.

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the N-th Lucas number.

Sample Input 1

5

Sample Output 1

11

L_0=2

L_1=1

L_2=L_0+L_1=3

L_3=L_1+L_2=4

L_4=L_2+L_3=7

L_5=L_3+L_4=11

Thus, the 5-th Lucas number is 11.

Sample Input 2

86

Sample Output 2

939587134549734843


(let ((lucas '(1 2))
      (n (read)))
  (defun lp ()
    (push (+ (first lucas) (second lucas)) lucas))
  (cond ((= 0 n) (princ 2))
        ((= 1 n) (princ 1))
        (t (loop :repeat (1- n) do(lp)) (princ (first lucas)))))


(defun bit-collection (n function)
  (loop :for k :from 0 :upto (expt 2 n)
          :thereis(funcall function (loop :for j :from 0 :upto (1- n) :collect (logbitp j k)))))
(defun parse-int-char (n)
  (cond ((char= #\0 n) 0)
        ((<= 49 (char-code n) 57) (- (char-code n) 48))
        (t nil)))
(let* ((lst (mapcar #'parse-int-char (concatenate 'list (read-line)) )))
  (defun f (ll)
    (if (= 7 (reduce #'+ (cons (car lst) (mapcar (lambda (x y) (if x y (- 0 y))) ll (cdr lst)))))
        (progn (princ (car lst)) (mapcar (lambda (x y) (princ (if x "+" "-")) (princ y)) ll (cdr lst)) (princ "=7"))))
  (bit-collection 3 #'f))

Score : 300 points

Problem Statement

Sitting in a station waiting room, Joisino is gazing at her train ticket.

The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).

In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with + or - so that the formula holds.

The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Constraints

0≤A,B,C,D≤9

All input values are integers.

It is guaranteed that there is a solution.

Input

Input is given from Standard Input in the following format:

ABCD

Output

Print the formula you made, including the part =7.

Use the signs + and -.

Do not print a space between a digit and a sign.

Sample Input 1

1222

Sample Output 1

1+2+2+2=7

This is the only valid solution.

Sample Input 2

0290

Sample Output 2

0-2+9+0=7

0 - 2 + 9 - 0 = 7 is also a valid solution.

Sample Input 3

3242

Sample Output 3

3+2+4-2=7


(let ((nm (mapcar (lambda (a) (parse-integer (string a))) (concatenate 'list (read-line))))
      (form '()))
  (setf nm (reverse nm))
  (loop :for k :from 0 :upto 7 :do
    (let ((kl (concatenate 'list (format nil "~3,'0B" k))))
      (setf form '())
      (push (first nm) form)
      (mapcar (lambda (x w)
                (if (char= x #\0)
                    (push w form)
                    (push (- w) form))) kl (cdr nm))
      form) :until (= 7 (reduce #'+ form)) :finally (format t "~D~@D~@D~@D=7" (nth 0 form) (nth 1 form) (nth 2 form) (nth 3 form))))


#include<iostream>
#include<string.h>

using namespace std;
const int inf = 1e7;
const int man = 1e2+10;
int maze[man][man];
int vis[man],dis[man];
int c[15][15];
int jishu[10];
int h,w;

int dij(int u)
{
	memset(vis,0,sizeof(vis));
	for(int i =1 ;i <= 9 ;i++)
	{
		dis[i] = c[u][i];
	}
	vis[u] = 1;
	for(int i =1 ;i < 9;i++)//必须是小于9，不能等于，debug了好久。。。
	{
		int minn = inf ,v = -1;
		for(int j =1 ;j <= 9 ;j++)
		{
			if(!vis[j]&&dis[j]<minn)
			{
				v = j;
				minn = dis[j];
			}
		}
		vis[v] = 1;
		for(int j = 1; j <= 9 ;j++)
		{
			dis[j] = min(dis[j],dis[v]+c[v][j]);
		}
	}
	return dis[1];
}

void solve()
{
	while(cin >> h >> w)
	{
		for(int i = 0; i < 10 ;i++)
		{
			for(int j = 0 ;j < 10 ;j++)
			{
				cin >> c[i][j];
			}
		}
		memset(jishu,0,sizeof(jishu));
		for(int i = 1 ; i <= w ;i++)
		{
			for(int j = 1; j<= h ;j++)
			{
				cin >> maze[i][j];
				if(maze[i][j] > 1)
				{
					jishu[maze[i][j]]++;
				}
			}
		}
		int ans = 0;
		//cout << dij(8) << endl;
		for(int i = 1 ; i <= 9 ;i++)
		{
			if(jishu[i])
			{
				int k = jishu[i];
				ans= ans +dij(i)*k;
			}
		//	cout << jishu[i] << endl;
		}
		cout << ans << endl;
	}

}

int main()
{
	solve();
	return 0;
}

Score : 400 points

Problem Statement

Joisino the magical girl has decided to turn every single digit that exists on this world into 1.

Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points).

She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive).

You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows:

If A_{i,j}≠-1, the square contains a digit A_{i,j}.

If A_{i,j}=-1, the square does not contain a digit.

Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

Constraints

1≤H,W≤200

1≤c_{i,j}≤10^3 (i≠j)

c_{i,j}=0 (i=j)

-1≤A_{i,j}≤9

All input values are integers.

There is at least one digit on the wall.

Input

Input is given from Standard Input in the following format:

H W
c_{0,0} ... c_{0,9}
:
c_{9,0} ... c_{9,9}
A_{1,1} ... A_{1,W}
:
A_{H,1} ... A_{H,W}

Output

Print the minimum total amount of MP required to turn every digit on the wall into 1 in the end.

Sample Input 1

2 4
0 9 9 9 9 9 9 9 9 9
9 0 9 9 9 9 9 9 9 9
9 9 0 9 9 9 9 9 9 9
9 9 9 0 9 9 9 9 9 9
9 9 9 9 0 9 9 9 9 2
9 9 9 9 9 0 9 9 9 9
9 9 9 9 9 9 0 9 9 9
9 9 9 9 9 9 9 0 9 9
9 9 9 9 2 9 9 9 0 9
9 2 9 9 9 9 9 9 9 0
-1 -1 -1 -1
8 1 1 8

Sample Output 1

12

To turn a single 8 into 1, it is optimal to first turn 8 into 4, then turn 4 into 9, and finally turn 9 into 1, costing 6 MP.

The wall contains two 8s, so the minimum total MP required is 6×2=12.

Sample Input 2

5 5
0 999 999 999 999 999 999 999 999 999
999 0 999 999 999 999 999 999 999 999
999 999 0 999 999 999 999 999 999 999
999 999 999 0 999 999 999 999 999 999
999 999 999 999 0 999 999 999 999 999
999 999 999 999 999 0 999 999 999 999
999 999 999 999 999 999 0 999 999 999
999 999 999 999 999 999 999 0 999 999
999 999 999 999 999 999 999 999 0 999
999 999 999 999 999 999 999 999 999 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1

Sample Output 2

0

Note that she may not need to change any digit.

Sample Input 3

3 5
0 4 3 6 2 7 2 5 3 3
4 0 5 3 7 5 3 7 2 7
5 7 0 7 2 9 3 2 9 1
3 6 2 0 2 4 6 4 2 3
3 5 7 4 0 6 9 7 6 7
9 8 5 2 2 0 4 7 6 5
5 4 6 3 2 3 0 5 4 3
3 6 2 3 4 2 4 0 8 9
4 6 5 4 3 5 3 2 0 8
2 1 3 4 5 7 8 6 4 0
3 5 2 6 1
2 5 3 2 1
6 9 2 5 6

Sample Output 3

47


(let* ((lst1 (list (car (concatenate 'list (symbol-name (read)))) (car (concatenate 'list  (symbol-name (read))))))
       (lst2 (sort (copy-list lst1) #'char-lessp)))

  (format t "~A~%"
          (if (= (length (remove-duplicates lst1)) 1)
            #\=
            (if (equal lst1 lst2)
              #\<
              #\>))))

Score : 100 points

Problem Statement

In programming, hexadecimal notation is often used.

In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters A, B, C, D, E and F are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.

In this problem, you are given two letters X and Y. Each X and Y is A, B, C, D, E or F.

When X and Y are seen as hexadecimal numbers, which is larger?

Constraints

Each X and Y is A, B, C, D, E or F.

Input

Input is given from Standard Input in the following format:

X Y

Output

If X is smaller, print <; if Y is smaller, print >; if they are equal, print =.

Sample Input 1

A B

Sample Output 1

<

10 < 11.

Sample Input 2

E C

Sample Output 2

>

14 > 12.

Sample Input 3

F F

Sample Output 3

=

15 = 15.


(let* ((a (read))
       (b (read))
       (c (read)))
  (setq a (- a c))
  (princ (floor a (+ b c))))

Score : 200 points

Problem Statement

We have a long seat of width X centimeters.
There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.

We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.

At most how many people can sit on the seat?

Constraints

All input values are integers.

1 \leq X, Y, Z \leq 10^5

Y+2Z \leq X

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

13 3 1

Sample Output 1

3

There is just enough room for three, as shown below:

Figure

Sample Input 2

12 3 1

Sample Output 2

2

Sample Input 3

100000 1 1

Sample Output 3

49999

Sample Input 4

64146 123 456

Sample Output 4

110

Sample Input 5

64145 123 456

Sample Output 5

109


Score : 300 points

Problem Statement

Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is YES or NO.

When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.

Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.

Now, he goes through the following process:

Submit the code.

Wait until the code finishes execution on all the cases.

If the code fails to correctly solve some of the M cases, submit it again.

Repeat until the code correctly solve all the cases in one submission.

Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).

Constraints

All input values are integers.

1 \leq N \leq 100

1 \leq M \leq {\rm min}(N, 5)

Input

Input is given from Standard Input in the following format:

N M

Output

Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.

Sample Input 1

1 1

Sample Output 1

3800

In this input, there is only one case. Takahashi will repeatedly submit the code that correctly solves this case with 1/2 probability in 1900 milliseconds.

The code will succeed in one attempt with 1/2 probability, in two attempts with 1/4 probability, and in three attempts with 1/8 probability, and so on.

Thus, the answer is 1900 \times 1/2 + (2 \times 1900) \times 1/4 + (3 \times 1900) \times 1/8 + ... = 3800.

Sample Input 2

10 2

Sample Output 2

18400

The code will take 1900 milliseconds in each of the 2 cases, and 100 milliseconds in each of the 10-2=8 cases. The probability of the code correctly solving all the cases is 1/2 \times 1/2 = 1/4.

Sample Input 3

100 5

Sample Output 3

608000


Score : 100 points

Problem Statement

You are given a grid with 2 rows and 3 columns of squares.
The color of the square at the i-th row and j-th column is represented by the character C_{ij}.

Write a program that prints YES if this grid remains the same when rotated 180 degrees, and prints NO otherwise.

Constraints

C_{i,j}(1 \leq i \leq 2, 1 \leq j \leq 3) is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

C_{11}C_{12}C_{13}
C_{21}C_{22}C_{23}

Output

Print YES if this grid remains the same when rotated 180 degrees; print NO otherwise.

Sample Input 1

pot
top

Sample Output 1

YES

This grid remains the same when rotated 180 degrees.

Sample Input 2

tab
bet

Sample Output 2

NO

This grid does not remain the same when rotated 180 degrees.

Sample Input 3

eye
eel

Sample Output 3

NO


Score : 200 points

Problem Statement

Find the largest square number not exceeding N. Here, a square number is an integer that can be represented as the square of an integer.

Constraints

1 \leq N \leq 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the largest square number not exceeding N.

Sample Input 1

10

Sample Output 1

9

10 is not square, but 9 = 3 × 3 is. Thus, we print 9.

Sample Input 2

81

Sample Output 2

81

Sample Input 3

271828182

Sample Output 3

271821169


(let ((n (read)))
  (loop for i from n downto 0
        for s = (sqrt i)
        when (= s (floor s)) do
        (format t "~A~%" i)
        (return)))


;; Utils (quoted from https://competitive12.blogspot.com/2020/03/common-lisp.html)

;; Read fixnum

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare (inline read-byte)
           #-swank (sb-kernel:ansi-stream in))
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (read-byte in nil 0))))
    (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) (the (integer 0 #.(floor most-positive-fixnum 10)) (* result 10))))
              (return (if minus (- result) result))))))))

;; fast read-line

(defmacro buffered-read-line (&optional (buffer-size 30) (in '*standard-input*) (term-char #\Space))
  (let ((buffer (gensym))
        (character (gensym))
        (idx (gensym)))
    `(let* ((,buffer (load-time-value (make-string ,buffer-size :element-type 'base-char))))
       (declare (simple-base-string ,buffer)
                (inline read-byte))
       (loop for ,character of-type base-char =
             ,(if (member :swank *features*)
                  `(read-char ,in nil #\Newline) ; on SLIME
                `(code-char (read-byte ,in nil #.(char-code #\Newline))))
             for ,idx from 0
             until (char= ,character #\Newline)
             do (setf (schar ,buffer ,idx) ,character)
             finally (when (< ,idx ,buffer-size)
                       (setf (schar ,buffer ,idx) ,term-char))
             (return (values ,buffer ,idx))))))

;;; invoke child process

#-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))))

;-------------------
; Write code here
;-------------------

(defun lower-bound (ok ng arr b-size)
  (labels ((judge (x b-size)
             (< (aref arr x) b-size))
           (bs (ok ng b-size)
             (if (< (abs (- ng ok)) 1)
                 ok
                 (let ((mid (floor (+ ng ok) 2)))
                   (if (judge mid b-size)
                       (bs mid ng b-size)
                       (bs ok mid b-size))))))
    (bs ok ng b-size)))

(defun upper-bound (ng ok arr b-size)
  (labels ((judge (x b-size)
             (> (aref arr x) b-size))
           (bs (ng ok b-size)
             (if (< (abs (- ng ok)) 1)
                 ok
                 (let ((mid (floor (+ ng ok) 2)))
                   (if (judge mid b-size)
                       (bs ng mid b-size)
                       (bs mid ok b-size))))))
    (bs ng ok b-size)))

(defun solve (n a b c &optional (res 0))
  (loop for j from 0 to n
        do (let ((i-cnt (1+ (lower-bound 0 n a (aref b j))))
                 (k-cnt (- n (upper-bound -1 (1- n) c (aref b j)))))
             (incf res (* i-cnt k-cnt)))
        finally
        (return res)))

(defun main ()
  (declare (inline sort))
  (let* ((n (read))
         (a (make-array n :initial-contents (read-from-string (concatenate 'string "(" (read-line) ")"))))
         (b (make-array n :initial-contents (read-from-string (concatenate 'string "(" (read-line) ")"))))
         (c (make-array n :initial-contents (read-from-string (concatenate 'string "(" (read-line) ")"))))
         )
    (princ (solve n (sort a #'<) (sort b #'<) (sort c #'<)))
    (fresh-line)))

(main)

Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

1 \leq C_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N
B_1 ... B_N
C_1 ... C_N

Output

Print the number of different altars that Ringo can build.

Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

3

The following three altars can be built:

Upper: 1-st part, Middle: 1-st part, Lower: 1-st part

Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part

Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

Sample Input 3

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

87


;; -*- 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
;;;
;;; In-place FFT
;;;
;;; Reference:
;;; http://www.prefield.com/algorithm/math/fft.html
;;; http://techtipshoge.blogspot.com/2011/08/fft4.html
;;;

(defconstant +dp-size+ (ash 1 18))
(deftype fft-float () 'single-float)

;; For FFT of fixed length, it is faster preparing the table of exp(i*theta).
(defun %make-cis-table (sign)
  (declare #.OPT
           ((integer -1 1) sign))
  (let* ((table (make-array #.(ash +dp-size+ -1) :element-type '(complex fft-float)))
         (theta (* sign #.(/ (coerce (* 2 pi) 'fft-float) +dp-size+))))
    (dotimes (i #.(ash +dp-size+ -1))
      (setf (aref table i) (cis (* i theta))))
    table))

(defconstant +cis+table+ (%make-cis-table 1))
(defconstant +cis-table+ (%make-cis-table -1))

(defun %general-dft-cached-cis! (f sign)
  (declare (optimize (speed 3) (safety 0) (debug 0))
           ((simple-array (complex fft-float) (#.+dp-size+)) f)
           ((integer -1 1) sign))
  (prog1 f
    (let ((table (if (= 1 sign) +cis+table+ +cis-table+)))
      (declare ((simple-array (complex fft-float) (#.(ash +dp-size+ -1))) table))
      (do ((m +dp-size+ (ash m -1))
           (shift 0 (+ shift 1)))
          ((= m 1))
        (declare ((integer 0 #.most-positive-fixnum) m shift))
        (let ((mh (ash m -1)))
          (dotimes (i mh)
            (do ((j i (+ j m)))
                ((>= j +dp-size+))
              (declare ((integer 0 #.most-positive-fixnum) j))
              (let* ((k (+ j mh))
                     (xt (- (aref f j) (aref f k)))
                     (cis-index (ash i shift)))
                (declare ((integer 0 #.most-positive-fixnum) k cis-index))
                (incf (aref f j) (aref f k))
                (setf (aref f k) (* (aref table cis-index) xt)))))))
      (let ((i 0))
        (declare ((integer 0 #.most-positive-fixnum) i))
        (loop for j from 1 below #.(- +dp-size+ 1)
              do (loop for k of-type (integer 0 #.most-positive-fixnum)
                          = #.(ash +dp-size+ -1) then (ash k -1)
                       while (> k (setq i (logxor i k))))
                 (when (< j i)
                   (rotatef (aref f i) (aref f j))))))))

(declaim (inline dft!))
(defun dft! (f)
  (declare #.OPT
           ((simple-array (complex fft-float) (#.+dp-size+)) f))
  (if (zerop (length f))
      f
      (%general-dft-cached-cis! f 1)))

(declaim (inline inverse-dft!))
(defun inverse-dft! (f)
  (declare #.OPT
           ((simple-array (complex fft-float) (#.+dp-size+)) f))
  (prog1 f
    (let ((n (length f)))
      (unless (zerop n)
        (let ((/n (/ (coerce n 'fft-float))))
          (%general-dft-cached-cis! f -1)
          (dotimes (i n)
            (setf (aref f i) (* (aref f i) /n))))))))

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

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

;; Body

(defun main ()
  (declare #.OPT)
  (let* ((k (read))
         (dp (make-array +dp-size+ :element-type '(complex fft-float) :initial-element #c(0f0 0f0)))
         (multiplier (make-array +dp-size+ :element-type '(complex fft-float) :initial-element #c(0f0 0f0))))
    (declare ((integer 0 100000) k)
             ((simple-array (complex fft-float) (#.+dp-size+)) dp multiplier))
    (loop for i = 1 then (rem (* i 10) k)
          while (= #c(0f0 0f0) (aref dp i))
          do (setf (aref dp i) #c(1f0 0f0)
                   (aref multiplier i) #c(1f0 0f0)))
    (dft! multiplier)
    (loop for i from 1 to 44
          do (unless (= #c(0f0 0f0) (aref dp 0))
               (println i)
               (return-from main))
             (dft! dp)
             (dotimes (i +dp-size+)
               (setf (aref dp i) (* (aref dp i) (aref multiplier i))))
             (inverse-dft! dp)
             (dotimes (i k)
               (if (or (> (realpart (aref dp i)) 0.5f0)
                       (> (realpart (aref dp (+ i k))) 0.5f0))
                   (setf (aref dp i) #c(1f0 0f0))
                   (setf (aref dp i) #c(0f0 0f0)))
               (setf (aref dp (+ i k)) #c(0f0 0f0)))
             (loop for i from (* 2 k) below +dp-size+
                   do (setf (aref dp i) #c(0f0 0f0))))
    (write-line "45")))

#-swank(main)

Score : 700 points

Problem Statement

Find the smallest possible sum of the digits in the decimal notation of a positive multiple of K.

Constraints

2 \leq K \leq 10^5

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print the smallest possible sum of the digits in the decimal notation of a positive multiple of K.

Sample Input 1

6

Sample Output 1

3

12=6×2 yields the smallest sum.

Sample Input 2

41

Sample Output 2

5

11111=41×271 yields the smallest sum.

Sample Input 3

79992

Sample Output 3

36


;; -*- 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
;;;
;;; Real FFT
;;;
;;; Reference:
;;; http://www.kurims.kyoto-u.ac.jp/~ooura/fftman/ftmn2_12.html#sec2_1_2
;;;

(deftype fft-float () 'single-float)

(declaim (inline power2-p))
(defun power2-p (x)
  "Checks if X is a power of 2."
  (zerop (logand x (- x 1))))

;; For FFT of fixed length, preparing the table of cos(i*theta) and sin
;; (i*theta) will be efficient.
(defun %make-trifunc-table (n)
  (declare (optimize (speed 3))
           ((integer 0 #.most-positive-fixnum) n))
  (assert (power2-p n))
  (let* ((cos-table (make-array (ash n -2) :element-type 'fft-float))
         (sin-table (make-array (ash n -2) :element-type 'fft-float))
         (theta (/ (coerce (* 2 pi) 'fft-float) n)))
    (dotimes (i (ash n -2))
      (setf (aref cos-table i) (cos (* i theta))
            (aref sin-table i) (sin (* i theta))))
    (values cos-table sin-table)))

(defparameter *cos-table* nil)
(defparameter *sin-table* nil)

(defmacro with-fixed-base (size &body body)
  "Makes FFT faster when the SIZE of target vectors is fixed in BODY. This macro
computes and holds the roots of unity for SIZE, which DFT! and INVERSE-DFT!
called in BODY automatically detects; they will signal an error when they
receive a vector of different size."
  (let ((s (gensym)))
    `(let ((,s ,size))
       (multiple-value-bind (*cos-table* *sin-table*) (%make-trifunc-table ,s)
         ,@body))))

(defun %dft-fixed-base! (f)
  (declare #.OPT
           ((simple-array fft-float (*)) f))
  (prog1 f
    (let* ((n (length f))
           (cos-table *cos-table*)
           (sin-table *sin-table*)
           (factor n))
      (declare ((integer 0 #.most-positive-fixnum) factor)
               ((simple-array fft-float (*)) cos-table sin-table))
      (assert (power2-p n))
      (assert (= (ash n -2) (length cos-table)))
      ;; bit-reverse ordering
      (let ((i 0))
        (declare ((integer 0 #.most-positive-fixnum) i))
        (loop for j from 1 below (- n 1)
              do (loop for k of-type (integer 0 #.most-positive-fixnum)
                          = (ash n -1) then (ash k -1)
                       while (> k (setq i (logxor i k))))
                 (when (< j i)
                   (rotatef (aref f i) (aref f j)))))
      (do* ((mh 1 m)
            (m (ash mh 1) (ash mh 1)))
           ((> m n))
        (declare ((integer 0 #.most-positive-fixnum) mh m))
        (let ((mq (ash mh -1)))
          (setq factor (ash factor -1))
          (do ((jr 0 (+ jr m)))
              ((>= jr n))
            (declare ((integer 0 #.most-positive-fixnum) jr))
            (let ((xreal (aref f (+ jr mh))))
              (setf (aref f (+ jr mh)) (- (aref f jr) xreal))
              (incf (aref f jr) xreal)))
          (do ((i 1 (+ i 1)))
              ((>= i mq))
            (declare ((integer 0 #.most-positive-fixnum) i))
            (let* ((index (the fixnum (* factor i)))
                   (wreal (aref cos-table index))
                   (wimag (- (aref sin-table index))))
              (do ((j 0 (+ j m)))
                  ((>= j n))
                (let* ((j+mh (+ j mh))
                       (j+m-i (- (+ j m) i))
                       (xreal (+ (* wreal (aref f (+ j+mh i)))
                                 (* wimag (aref f j+m-i))))
                       (ximag (- (* wreal (aref f j+m-i))
                                 (* wimag (aref f (+ j+mh i))))))
                  (declare ((integer 0 #.most-positive-fixnum) j+mh j+m-i))
                  (setf (aref f (+ j+mh i))
                        (+ (- (aref f (- j+mh i))) ximag))
                  (setf (aref f j+m-i)
                        (+ (aref f (- j+mh i)) ximag))
                  (setf (aref f (- j+mh i))
                        (+ (aref f (+ j i)) (- xreal)))
                  (incf (aref f (+ j i)) xreal))))))))))

(defun %inverse-dft-fixed-base! (f)
  (declare #.OPT
           ((simple-array fft-float (*)) f))
  (prog1 f
    (let* ((n (length f))
           (cos-table *cos-table*)
           (sin-table *sin-table*)
           (factor 1))
      (declare ((integer 0 #.most-positive-fixnum) factor)
               ((simple-array fft-float (*)) cos-table sin-table))
      (assert (power2-p n))
      (assert (= (ash n -2) (length cos-table)))
      (setf (aref f 0)
            (/ (aref f 0) 2))
      (setf (aref f (ash n -1))
            (/ (aref f (ash n -1)) 2))
      (do* ((m n mh)
            (mh (ash m -1) (ash m -1)))
           ((zerop mh))
        (declare ((integer 0 #.most-positive-fixnum) m mh))
        (let ((mq (ash mh -1)))
          (do ((jr 0 (+ jr m)))
              ((>= jr n))
            (declare ((integer 0 #.most-positive-fixnum) jr))
            (let ((xreal (- (aref f jr) (aref f (+ jr mh)))))
              (incf (aref f jr) (aref f (+ jr mh)))
              (setf (aref f (+ jr mh)) xreal)))
          (do ((i 1 (+ i 1)))
              ((>= i mq))
            (let* ((index (the fixnum (* factor i)))
                   (wreal (aref cos-table index))
                   (wimag (aref sin-table index)))
              (do ((j 0 (+ j m)))
                  ((>= j n))
                (let* ((j+mh (+ j mh))
                       (j+m-i (- (+ j m) i))
                       (xreal (- (aref f (+ j i)) (aref f (- j+mh i))))
                       (ximag (+ (aref f j+m-i) (aref f (+ j+mh i)))))
                  (declare ((integer 0 #.most-positive-fixnum) j+mh j+m-i))
                  (incf (aref f (+ j i)) (aref f (- j+mh i)))
                  (setf (aref f (- j+mh i))
                        (- (aref f j+m-i) (aref f (+ j+mh i))))
                  (setf (aref f (+ j+mh i))
                        (+ (* wreal xreal) (* wimag ximag)))
                  (setf (aref f j+m-i)
                        (- (* wreal ximag) (* wimag xreal))))))))
        (setq factor (ash factor 1)))
      ;; bit-reverse ordering
      (let ((i 0))
        (declare ((integer 0 #.most-positive-fixnum) i))
        (loop for j from 1 below (- n 1)
              do (loop for k of-type (integer 0 #.most-positive-fixnum)
                          = (ash n -1) then (ash k -1)
                       while (> k (setq i (logxor i k))))
                 (when (< j i)
                   (rotatef (aref f i) (aref f j))))))))

(declaim (inline dft!))
(defun dft! (f)
  (declare ((simple-array fft-float (*)) f))
  (if (zerop (length f))
      f
      (%dft-fixed-base! f)))

(declaim (inline inverse-dft!))
(defun inverse-dft! (f)
  (declare ((simple-array fft-float (*)) f))
  (prog1 f
    (let ((n (length f)))
      (unless (zerop n)
        (let ((factor (* 2 (/ (coerce n 'fft-float)))))
          (%inverse-dft-fixed-base! f)
          (dotimes (i n)
            (setf (aref f i) (* (aref f i) factor))))))))

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

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

;; Body
(defconstant +dp-size+ (ash 1 18))

(defun main ()
  (declare #.OPT)
  (let* ((k (read))
         (dp (make-array +dp-size+ :element-type 'fft-float :initial-element 0f0))
         (multiplier (make-array +dp-size+ :element-type 'fft-float :initial-element 0f0)))
    (declare ((integer 0 100000) k)
             ((simple-array fft-float (#.+dp-size+)) dp multiplier))
    (loop for i = 1 then (rem (* i 10) k)
          while (= 0f0 (aref dp i))
          do (setf (aref dp i) 1f0
                   (aref multiplier i) 1f0))
    (with-fixed-base +dp-size+
      (dft! multiplier)
      (loop for i from 1 to 45
            do (when (= 1f0 (aref dp 0))
                 (println i)
                 (return-from main))
               (dft! dp)
               (setf (aref dp 0)
                     (* (aref dp 0) (aref multiplier 0)))
               (setf (aref dp (ash +dp-size+ -1))
                     (* (aref dp (ash +dp-size+ -1)) (aref multiplier (ash +dp-size+ -1))))
               (loop for i from 1 below (ash +dp-size+ -1)
                     for value1 = (- (* (aref dp i) (aref multiplier i))
                                     (* (aref dp (- +dp-size+ i)) (aref multiplier (- +dp-size+ i))))
                     for value2 = (+ (* (aref dp i) (aref multiplier (- +dp-size+ i)))
                                     (* (aref dp (- +dp-size+ i)) (aref multiplier i)))
                     do (setf (aref dp i) value1
                              (aref dp (- +dp-size+ i)) value2))
               (inverse-dft! dp)
               (dotimes (i k)
                 (if (or (> (aref dp i) 0.5f0)
                         (> (aref dp (+ i k)) 0.5f0))
                     (setf (aref dp i) 1f0)
                     (setf (aref dp i) 0f0))
                 (setf (aref dp (+ i k)) 0f0))
               (loop for i from (* 2 k) below +dp-size+
                     do (setf (aref dp i) 0f0))))))

#-swank(main)


(let ((a (read))
      (b (read)))
  (format t "~a~%" (floor (- (* b 2) a))))

Score : 100 points

Problem Statement

Takahashi is a user of a site that hosts programming contests.

When a user competes in a contest, the rating of the user (not necessarily an integer) changes according to the performance of the user, as follows:

Let the current rating of the user be a.

Suppose that the performance of the user in the contest is b.

Then, the new rating of the user will be the avarage of a and b.

For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.

Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.

Find the performance required to achieve it.

Constraints

0 \leq R, G \leq 4500

All input values are integers.

Input

Input is given from Standard Input in the following format:

R
G

Output

Print the performance required to achieve the objective.

Sample Input 1

2002
2017

Sample Output 1

2032

Takahashi's current rating is 2002.

If his performance in the contest is 2032, his rating will be the average of 2002 and 2032, which is equal to the desired rating, 2017.

Sample Input 2

4500
0

Sample Output 2

-4500

Although the current and desired ratings are between 0 and 4500, the performance of a user can be below 0.


(defun f (x y &optional (ans 1))
  (if (< x 1)
    ans
    (f (1- x) y (min (* 2 ans) (+ y ans)))))

(let ((n (read))
      (k (read)))

  (format t "~A~%"
          (f n k)))

Score : 200 points

Problem Statement

Square1001 has seen an electric bulletin board displaying the integer 1.
He can perform the following operations A and B to change this value:

Operation A: The displayed value is doubled.

Operation B: The displayed value increases by K.

Square1001 needs to perform these operations N times in total.
Find the minimum possible value displayed in the board after N operations.

Constraints

1 \leq N, K \leq 10

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the minimum possible value displayed in the board after N operations.

Sample Input 1

4
3

Sample Output 1

10

The value will be minimized when the operations are performed in the following order: A, A, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 7 → 10.

Sample Input 2

10
10

Sample Output 2

76

The value will be minimized when the operations are performed in the following order: A, A, A, A, B, B, B, B, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 8 → 16 → 26 → 36 → 46 → 56 → 66 → 76.

By the way, this contest is AtCoder Beginner Contest 076.


(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) (the fixnum (* result 10))))
              (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))
         (ts (make-array n :element-type 'uint16))
         (cumul-ts (make-array (+ 1 n) :element-type 'uint16 :initial-element 0))
         (vs (make-array (+ n 1) :element-type 'uint16)))
    (dotimes (i n) (setf (aref ts i) (* 2 (read-fixnum))))
    (dotimes (i n) (setf (aref vs i) (* 2 (read-fixnum))))
    (loop for i from n downto 1
          do (when (> (aref vs (- i 1)) (aref vs i))
               (setf (aref vs (- i 1))
                     (min (aref vs (- i 1))
                          (+ (aref vs i) (aref ts (- i 1)))))))
    (dotimes (i n)
      (setf (aref cumul-ts (+ i 1))
            (+ (aref cumul-ts i) (aref ts i))))
    (let* ((total-time (reduce #'+ ts))
           (next-points (make-array total-time :element-type 'uint16))
           (prev-points (make-array total-time :element-type 'uint16))
           (time-vs (make-array total-time :element-type 'uint16))
           (time-next-vs (make-array total-time :element-type 'uint16))
           (time-prev-vs (make-array total-time :element-type 'uint16))
           (timeline (make-array (+ total-time 1) :element-type 'uint16 :initial-element 0)))
      (dotimes (i n)
        (loop for time from (aref cumul-ts i) below (aref cumul-ts (+ i 1))
              do (setf (aref next-points time) (aref cumul-ts (+ i 1))
                       (aref prev-points time) (aref cumul-ts i)
                       (aref time-vs time) (aref vs i)
                       (aref time-next-vs time) (aref vs (+ i 1))
                       (aref time-prev-vs time) (if (zerop i) 0 (aref vs (- i 1))))))
      (dotimes (time total-time)
        (setf (aref timeline time)
              (min (aref time-vs time)
                   (+ (aref time-next-vs time)
                      (- (aref next-points time) time))
                   (+ (aref time-prev-vs time)
                      (- time (aref prev-points time))))))
      (println (/ (loop for time from 0 below total-time
                        sum (+ (aref timeline time) (aref timeline (+ 1 time))))
                  8d0)))))

#-swank(main)

Score : 400 points

Problem Statement

In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.

In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.

According to the specifications of the trains, the acceleration of a train must be always within ±1m/s^2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

1 \leq N \leq 100

1 \leq t_i \leq 200

1 \leq v_i \leq 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N

Output

Print the maximum possible that a train can cover in the run.

Output is considered correct if its absolute difference from the judge's output is at most 10^{-3}.

Sample Input 1

1
100
30

Sample Output 1

2100.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 30 seconds, it accelerates at a rate of 1m/s^2, covering 450 meters.

In the subsequent 40 seconds, it maintains the velocity of 30m/s, covering 1200 meters.

In the last 30 seconds, it decelerates at the acceleration of -1m/s^2, covering 450 meters.

The total distance covered is 450 + 1200 + 450 = 2100 meters.

Sample Input 2

2
60 50
34 38

Sample Output 2

2632.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 34 seconds, it accelerates at a rate of 1m/s^2, covering 578 meters.

In the subsequent 26 seconds, it maintains the velocity of 34m/s, covering 884 meters.

In the subsequent 4 seconds, it accelerates at a rate of 1m/s^2, covering 144 meters.

In the subsequent 8 seconds, it maintains the velocity of 38m/s, covering 304 meters.

In the last 38 seconds, it decelerates at the acceleration of -1m/s^2, covering 722 meters.

The total distance covered is 578 + 884 + 144 + 304 + 722 = 2632 meters.

Sample Input 3

3
12 14 2
6 2 7

Sample Output 3

76.000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 6 seconds, it accelerates at a rate of 1m/s^2, covering 18 meters.

In the subsequent 2 seconds, it maintains the velocity of 6m/s, covering 12 meters.

In the subsequent 4 seconds, it decelerates at the acceleration of -1m/s^2, covering 16 meters.

In the subsequent 14 seconds, it maintains the velocity of 2m/s, covering 28 meters.

In the last 2 seconds, it decelerates at the acceleration of -1m/s^2, covering 2 meters.

The total distance covered is 18 + 12 + 16 + 28 + 2 = 76 meters.

Sample Input 4

1
9
10

Sample Output 4

20.250000000000000000

The maximum distance is achieved when a train runs as follows:

In the first 4.5 seconds, it accelerates at a rate of 1m/s^2, covering 10.125 meters.

In the last 4.5 seconds, it decelerates at the acceleration of -1m/s^2, covering 10.125 meters.

The total distance covered is 10.125 + 10.125 = 20.25 meters.

Sample Input 5

10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70

Sample Output 5

20291.000000000000


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

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

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

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

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

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun main ()
  (let* ((n (read))
         (res 1))
    (loop repeat n
          for a = (read)
          do (mulf res (if (evenp a) 2 1)))
    (println (- (expt 3 n) res))))

#-swank (main)

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

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

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

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

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

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

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

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

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2
2 3
"
    "7
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "3
3 3 3
"
    "26
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "1
100
"
    "1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "10
90 52 56 71 44 8 13 30 57 84
"
    "58921
")))

Score : 200 points

Problem Statement

We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are similar when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N).

In particular, any integer sequence is similar to itself.

You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N.

How many integer sequences b_1, b_2, ..., b_N are there such that b_1, b_2, ..., b_N is similar to A and the product of all elements, b_1 b_2 ... b_N, is even?

Constraints

1 \leq N \leq 10

1 \leq A_i \leq 100

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the number of integer sequences that satisfy the condition.

Sample Input 1

2
2 3

Sample Output 1

7

There are seven integer sequences that satisfy the condition:

1, 2

1, 4

2, 2

2, 3

2, 4

3, 2

3, 4

Sample Input 2

3
3 3 3

Sample Output 2

26

Sample Input 3

1
100

Sample Output 3

1

Sample Input 4

10
90 52 56 71 44 8 13 30 57 84

Sample Output 4

58921


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

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

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

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

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

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun solve (n s)
  (declare #.OPT
           (uint31 n)
           ((simple-array uint32 (*)) s))
  (let* ((cumul (make-array (+ n 1) :element-type 'uint32 :initial-element 0))
         (highest (make-hash-table :size n :test #'eq)))
    (dotimes (i n)
      (setf (aref cumul (+ i 1))
            (logxor (aref cumul i) (aref s i))))
    (dotimes (i (+ n 1))
      (push i (gethash (aref cumul i) highest)))
    (when (zerop (aref cumul n))
      (return-from solve 1))
    (let ((dp (make-array (+ n 1) :element-type 'uint32 :initial-element #xffffffff)))
      (setf (aref dp 0) 0)
      (dotimes (x n)
        ;; a to z
        (unless (= #xffffffff (aref dp x))
          (dotimes (c 26)
            (let* ((mask (ash 1 c))
                   (target (logxor (aref cumul x) mask))
                   (nexts (gethash target highest)))
              (loop repeat 3
                    for next of-type (or null uint32) in nexts
                    while (and next (> next x))
                    do (minf (aref dp next) (+ 1 (aref dp x))))))))
      (assert (/= #xffffffff (aref dp n)))
      (aref dp n))))

(defun main ()
  (let* ((orig-s (read-line))
         (n (length orig-s))
         (s (make-array n :element-type 'uint32)))
    (dotimes (i n)
      (let ((c (aref orig-s i)))
        (setf (aref s i) (ash 1 (- (char-code c) 97)))))
    (println (min (solve n s)
                  (solve n (reverse 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 "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 "aabxyyzz
"
    "2
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "byebye
"
    "1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "abcdefghijklmnopqrstuvwxyz
"
    "26
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "abcabcxabcx
"
    "3
")))

Score : 700 points

Problem Statement

We have a string s consisting of lowercase English letters.
Snuke is partitioning s into some number of non-empty substrings.
Let the subtrings obtained be s_1, s_2, ..., s_N from left to right. (Here, s = s_1 + s_2 + ... + s_N holds.)
Snuke wants to satisfy the following condition:

For each i (1 \leq i \leq N), it is possible to permute the characters in s_i and obtain a palindrome.

Find the minimum possible value of N when the partition satisfies the condition.

Constraints

1 \leq |s| \leq 2 \times 10^5

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum possible value of N when the partition satisfies the condition.

Sample Input 1

aabxyyzz

Sample Output 1

2

The solution is to partition s as aabxyyzz = aab + xyyzz.
Here, aab can be permuted to form a palindrome aba, and xyyzz can be permuted to form a palindrome zyxyz.

Sample Input 2

byebye

Sample Output 2

1

byebye can be permuted to form a palindrome byeeyb.

Sample Input 3

abcdefghijklmnopqrstuvwxyz

Sample Output 3

26

Sample Input 4

abcabcxabcx

Sample Output 4

3

The solution is to partition s as abcabcxabcx = a + b + cabcxabcx.


Score : 100 points

Problem Statement

You are given three integers, A, B and C.

Among them, two are the same, but the remaining one is different from the rest.

For example, when A=5,B=7,C=5, A and C are the same, but B is different.

Find the one that is different from the rest among the given three integers.

Constraints

-100 \leq A,B,C \leq 100

A, B and C are integers.

The input satisfies the condition in the statement.

Input

Input is given from Standard Input in the following format:

A B C

Output

Among A, B and C, print the integer that is different from the rest.

Sample Input 1

5 7 5

Sample Output 1

7

This is the same case as the one in the statement.

Sample Input 2

1 1 7

Sample Output 2

7

In this case, C is the one we seek.

Sample Input 3

-100 100 100

Sample Output 3

-100


(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
(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 ()
  (let* ((s (read-line)))
    (write-line s *standard-output* :end (- (length s) 8))))

#-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 "CODEFESTIVAL
"
    "CODE
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "CODEFESTIVALFESTIVAL
"
    "CODEFESTIVAL
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "YAKINIKUFESTIVAL
"
    "YAKINIKU
")))

Score : 100 points

Problem Statement

Rng is going to a festival.

The name of the festival is given to you as a string S, which ends with FESTIVAL, from input. Answer the question: "Rng is going to a festival of what?" Output the answer.

Here, assume that the name of "a festival of s" is a string obtained by appending FESTIVAL to the end of s.
For example, CODEFESTIVAL is a festival of CODE.

Constraints

9 \leq |S| \leq 50

S consists of uppercase English letters.

S ends with FESTIVAL.

Input

Input is given from Standard Input in the following format:

S

Output

Print the answer to the question: "Rng is going to a festival of what?"

Sample Input 1

CODEFESTIVAL

Sample Output 1

CODE

This is the same as the example in the statement.

Sample Input 2

CODEFESTIVALFESTIVAL

Sample Output 2

CODEFESTIVAL

This string is obtained by appending FESTIVAL to the end of CODEFESTIVAL, so it is a festival of CODEFESTIVAL.

Sample Input 3

YAKINIKUFESTIVAL

Sample Output 3

YAKINIKU


(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"
  (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))))))))

(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 ()
  (let* ((n (read))
         (table (make-hash-table :test #'eq)))
    (dotimes (_ n)
      (let ((d (read-fixnum)))
        (if (gethash d table)
            (incf (the fixnum (gethash d table)))
            (setf (gethash d table) 1))))
    (let ((m (read)))
      (dotimes (_ m)
        (let ((d (read-fixnum)))
          (unless (gethash d table)
            (write-line "NO")
            (return-from main))
          (if (eql 1 (gethash d table))
              (remhash d table)
              (decf (the fixnum (gethash d table))))))
      (write-line "YES"))))

#-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 "5
3 1 4 1 5
3
5 4 3
"
    "YES
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "7
100 200 500 700 1200 1600 2000
6
100 200 500 700 1600 1600
"
    "NO
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "1
800
5
100 100 100 100 100
"
    "NO
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "15
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
9
5 4 3 2 1 2 3 4 5
"
    "YES
")))

Score : 200 points

Problem Statement

Rng is preparing a problem set for a qualification round of CODEFESTIVAL.

He has N candidates of problems. The difficulty of the i-th candidate is D_i.

There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems.

Determine whether Rng can complete the problem set without creating new candidates of problems.

Constraints

1 \leq N \leq 200,000

1 \leq D_i \leq 10^9

1 \leq M \leq 200,000

1 \leq T_i \leq 10^9

All numbers in the input are integers.

Partial Score

100 points will be awarded for passing the test set satisfying N \leq 100 and M \leq 100.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N
M
T_1 T_2 ... T_M

Output

Print YES if Rng can complete the problem set without creating new candidates of problems; print NO if he cannot.

Sample Input 1

5
3 1 4 1 5
3
5 4 3

Sample Output 1

YES

Sample Input 2

7
100 200 500 700 1200 1600 2000
6
100 200 500 700 1600 1600

Sample Output 2

NO

Not enough 1600s.

Sample Input 3

1
800
5
100 100 100 100 100

Sample Output 3

NO

Sample Input 4

15
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
9
5 4 3 2 1 2 3 4 5

Sample Output 4

YES