(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
(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))

(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
;;;

(defconstant +inf+ 10001)
(defconstant +nan+ -1)
(defun main ()
  (let* ((n (read))
         (dp (make-array (list (+ n 1) 10002) :element-type 'fixnum :initial-element +nan+))
         (towers (make-array n :element-type 'list)))
    (dotimes (i n)
      (setf (aref towers i) (list (read) (read) (read))))
    (setq towers (sort towers #'> :key (lambda (node) (+ (first node) (second node)))))
    (setf (aref dp 0 +inf+) 0)
    (dotimes (x n)
      (destructuring-bind (w s v) (aref towers x)
        (declare (uint31 w s v))
        (loop for y from 0 to 10000
              unless (= (aref dp x y) +nan+)
              do (maxf (aref dp (+ x 1) y) (aref dp x y))
                 (let ((next-y (min s (- y w))))
                   (when (>= next-y 0)
                     (maxf (aref dp (+ x 1) next-y)
                           (+ v (aref dp x y))))))
        (maxf (aref dp (+ x 1) +inf+)
              (aref dp x +inf+))
        (maxf (aref dp (+ x 1) s)
              (+ v (aref dp x +inf+)))))
    (println (loop for y from 0 below 10002
                   maximize (aref dp n y)))))

#-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 "3
2 2 20
2 1 30
3 1 40
"
    "50
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "4
1 2 10
3 1 10
2 4 10
1 6 10
"
    "40
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "5
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
"
    "5000000000
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "8
9 5 7
6 2 7
5 7 3
7 8 8
1 9 6
3 3 3
4 1 7
4 5 5
"
    "22
")))

Score : 100 points

Problem Statement

There are N blocks, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), Block i has a weight of w_i, a solidness of s_i and a value of v_i.

Taro has decided to build a tower by choosing some of the N blocks and stacking them vertically in some order.
Here, the tower must satisfy the following condition:

For each Block i contained in the tower, the sum of the weights of the blocks stacked above it is not greater than s_i.

Find the maximum possible sum of the values of the blocks contained in the tower.

Constraints

All values in input are integers.

1 \leq N \leq 10^3

1 \leq w_i, s_i \leq 10^4

1 \leq v_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
w_1 s_1 v_1
w_2 s_2 v_2
:
w_N s_N v_N

Output

Print the maximum possible sum of the values of the blocks contained in the tower.

Sample Input 1

3
2 2 20
2 1 30
3 1 40

Sample Output 1

50

If Blocks 2, 1 are stacked in this order from top to bottom, this tower will satisfy the condition, with the total value of 30 + 20 = 50.

Sample Input 2

4
1 2 10
3 1 10
2 4 10
1 6 10

Sample Output 2

40

Blocks 1, 2, 3, 4 should be stacked in this order from top to bottom.

Sample Input 3

5
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000

Sample Output 3

5000000000

The answer may not fit into a 32-bit integer type.

Sample Input 4

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

Sample Output 4

22

We should, for example, stack Blocks 5, 6, 8, 4 in this order from top to bottom.


(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
;;;
;;; Convex Hull Trick for monotone slopes
;;;

(deftype cht-element-type () 'fixnum)

(define-condition cht-empty-error (error)
  ((cht :initarg :cht :accessor cht-empty-error-cht))
  (:report (lambda (condition stream)
             (format stream
                     "Attempted to get a value on an empty CHT ~W"
                     (cht-empty-error-cht condition)))))

(define-condition cht-full-error (error)
  ((cht :initarg :cht :accessor cht-full-error-cht))
  (:report (lambda (condition stream)
             (format stream
                     "Attempted to push a value on a full CHT ~W"
                     (cht-full-error-cht condition)))))

(defstruct (convex-hull-trick
            (:constructor make-cht
                (max-length
                 &optional (minimum t)
                 &aux
                 (slopes (make-array max-length :element-type 'cht-element-type))
                 (intercepts (make-array max-length :element-type 'cht-element-type))))
            (:conc-name %cht-)
            (:copier nil))
  (slopes nil :type (simple-array cht-element-type (*)))
  (intercepts nil :type (simple-array cht-element-type (*)))
  (minimum t :type boolean)
  (start 0 :type (integer 0 #.most-positive-fixnum))
  (length 0 :type (integer 0 #.most-positive-fixnum))
  (max-length 0 :type (integer 0 #.most-positive-fixnum)))

;; four basic operations on deque
(declaim (inline %cht-pop-back))
(defun %cht-pop-back (cht)
  (decf (%cht-length cht)))

(declaim (inline %cht-pop-front))
(defun %cht-pop-front (cht)
  (decf (%cht-length cht))
  (incf (%cht-start cht))
  (when (= (%cht-start cht) (%cht-max-length cht))
    (setf (%cht-start cht) 0)))

(declaim (inline %cht-push-back))
(defun %cht-push-back (cht slope intercept)
  (let ((pos (+ (%cht-start cht) (%cht-length cht))))
    (declare ((integer 0 #.most-positive-fixnum) pos))
    (when (>= pos (%cht-max-length cht))
      (decf pos (%cht-max-length cht)))
    (setf (aref (%cht-slopes cht) pos) slope
          (aref (%cht-intercepts cht) pos) intercept)
    (incf (%cht-length cht))))

(declaim (inline %cht-push-front))
(defun %cht-push-front (cht slope intercept)
  (let ((new-start (- (%cht-start cht) 1)))
    (when (= -1 new-start)
      (incf new-start (%cht-max-length cht)))
    (setf (aref (%cht-slopes cht) new-start) slope
          (aref (%cht-intercepts cht) new-start) intercept)
    (setf (%cht-start cht) new-start)
    (incf (%cht-length cht))))

(declaim (inline %removable-p))
(defun %removable-p (slope1 intercept1 slope2 intercept2 slope3 intercept3)
  "Returns true iff the **second** line is removable."
  (>= (* (- intercept3 intercept2)
         (- slope2 slope1))
      (* (- intercept2 intercept1)
         (- slope3 slope2))))

;; NOTE: The slopes of lines newly added to CHT must be largest or smallest
;; ever.
(defun cht-push (cht slope intercept)
  "Adds a new line to CHT."
  (declare (optimize (speed 3))
           (cht-element-type slope intercept))
  (when (= (%cht-length cht) (%cht-max-length cht))
    (error 'cht-full-error :cht cht))
  (unless (%cht-minimum cht)
    (setq slope (- slope)
          intercept (- intercept)))
  (let ((slopes (%cht-slopes cht))
        (intercepts (%cht-intercepts cht)))
    (labels ((ref (i)
               (let ((pos (+ (%cht-start cht) i)))
                 (declare ((integer 0 #.most-positive-fixnum) pos))
                 (when (>= pos (%cht-max-length cht))
                   (decf pos (%cht-max-length cht)))
                 (values (aref slopes pos) (aref intercepts pos)))))
      (cond ((zerop (%cht-length cht))
             (%cht-push-front cht slope intercept))
            ((>= slope (aref slopes (%cht-start cht)))
             ;; push the line to the front if SLOPE is larger than that of the head.
             (loop for start = (%cht-start cht)
                   while (and (>= (%cht-length cht) 2)
                              (let ((slope+1 (aref slopes start))
                                    (intercept+1 (aref intercepts start)))
                                (multiple-value-bind (slope+2 intercept+2) (ref 1)
                                  (declare (cht-element-type slope+1 intercept+1 slope+2 intercept+2))
                                  (%removable-p slope intercept
                                                slope+1 intercept+1
                                                slope+2 intercept+2))))
                   do (%cht-pop-front cht)
                   finally (%cht-push-front cht slope intercept)))
            (t
             ;; push the line to the end if SLOPE is smaller than that of the tail.
             ;; TODO: assert it.
             (loop for offset = (%cht-length cht)
                   while (and (>= offset 2)
                              (multiple-value-bind (slope-2 intercept-2) (ref (- offset 2))
                                (multiple-value-bind (slope-1 intercept-1) (ref (- offset 1))
                                  (declare (cht-element-type slope-2 intercept-2 slope-1 intercept-1))
                                  (%removable-p slope-2 intercept-2
                                                slope-1 intercept-1
                                                slope intercept))))
                   do (%cht-pop-back cht)
                   finally (%cht-push-back cht slope intercept))))
      cht)))

(declaim (inline cht-get))
(defun cht-get (cht x)
  "Returns the minimum (maximum) value at X. The time complexity is O(log(n))."
  (when (zerop (%cht-length cht))
    (error 'cht-empty-error :cht cht))
  (let ((ng -1)
        (ok (- (%cht-length cht) 1))
        (slopes (%cht-slopes cht))
        (intercepts (%cht-intercepts cht)))
    (declare ((integer -1 (#.array-total-size-limit)) ng ok))
    (labels ((calc (i)
               (let ((pos (+ (%cht-start cht) i)))
                 (declare ((integer 0 #.most-positive-fixnum) pos))
                 (when (>= pos (%cht-max-length cht))
                   (decf pos (%cht-max-length cht)))
                 (+ (* x (aref slopes pos))
                    (aref intercepts pos)))))
      (loop
        (when (<= (- ok ng) 1)
          (return
            (if (%cht-minimum cht)
                (calc ok)
                (- (calc ok)))))
        (let ((mid (ash (+ ng ok) -1)))
          (if (< (calc mid) (calc (+ mid 1)))
              (setq ok mid)
              (setq ng mid)))))))

(declaim (inline cht-increasing-get))
(defun cht-increasing-get (cht x)
  "Returns the minimum (maximum) value at X. The time complexity is O(1), though
X must be larger than the one given at the previous call of this function."
  (when (zerop (%cht-length cht))
    (error 'cht-empty-error :cht cht))
  (let ((slopes (%cht-slopes cht))
        (intercepts (%cht-intercepts cht)))
    (labels ((calc (slope intercept)
               (declare (cht-element-type slope intercept))
               (+ (* x slope) intercept)))
      (loop while (and (>= (%cht-length cht) 2)
                       (let* ((pos (%cht-start cht))
                              (slope0 (aref slopes pos))
                              (intercept0 (aref intercepts pos)))
                         (incf pos)
                         (when (= pos (%cht-max-length cht))
                           (setq pos 0))
                         (let ((slope1 (aref slopes pos))
                               (intercept1 (aref intercepts pos)))
                           (>= (calc slope0 intercept0)
                               (calc slope1 intercept1)))))
            do (%cht-pop-front cht))
      (let ((start (%cht-start cht)))
        (if (%cht-minimum cht)
            (calc (aref slopes start) (aref intercepts start))
            (- (calc (aref slopes start) (aref intercepts start))))))))

(declaim (inline cht-decreasing-get))
(defun cht-decreasing-get (cht x)
  "Returns the minimum (maximum) value at X. The time complexity is O(1), though
X must be smaller than the one given at the previous call of this function."
  (when (zerop (%cht-length cht))
    (error 'cht-empty-error :cht cht))
  (let ((slopes (%cht-slopes cht))
        (intercepts (%cht-intercepts cht)))
    (labels ((calc (slope intercept)
               (declare (cht-element-type slope intercept))
               (+ (* x slope) intercept))
             (get-last-idx ()
               (let ((idx (+ (%cht-start cht) (%cht-length cht) -1)))
                 (if (>= idx (%cht-max-length cht))
                     (- idx (%cht-max-length cht))
                     idx))))
      (loop while (and (>= (%cht-length cht) 2)
                       (let* ((pos (get-last-idx))
                              (slope-1 (aref slopes pos))
                              (intercept-1 (aref intercepts pos)))
                         (decf pos)
                         (when (< pos 0)
                           (incf pos (%cht-max-length cht)))
                         (let ((slope-2 (aref slopes pos))
                               (intercept-2 (aref intercepts pos)))
                           (>= (calc slope-1 intercept-1)
                               (calc slope-2 intercept-2)))))
            do (%cht-pop-back cht))
      (let ((end-1 (get-last-idx)))
        (if (%cht-minimum cht)
            (calc (aref slopes end-1) (aref intercepts end-1))
            (- (calc (aref slopes end-1) (aref intercepts end-1))))))))

(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 ()
  (declare #.OPT)
  (let* ((n (read))
         (c (read))
         (hs (make-array n :element-type 'uint31 :initial-element 0))
         (dp (make-array n :element-type 'uint62 :initial-element 0))
         (cht (make-cht n)))
    (dotimes (i n)
      (setf (aref hs i) (read-fixnum)))
    (cht-push cht (* -2 (aref hs 0)) (expt (aref hs 0) 2))
    (loop for x from 1 below n
          for h = (aref hs x)
          for value = (+ (* h h) (cht-get cht h) c)
          do (setf (aref dp x) value)
             (cht-push cht (* -2 h) (+ (* h h) value)))
    (println (aref dp (- n 1)))))

#-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 "5 6
1 2 3 4 5
"
    "20
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "2 1000000000000
500000 1000000
"
    "1250000000000
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "8 5
1 3 4 5 10 11 12 13
"
    "62
")))

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

Score : 100 points

Problem Statement

There are N stones, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), the height of Stone i is h_i.
Here, h_1 < h_2 < \cdots < h_N holds.

There is a frog who is initially on Stone 1.
He will repeat the following action some number of times to reach Stone N:

If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, N. Here, a cost of (h_j - h_i)^2 + C is incurred, where j is the stone to land on.

Find the minimum possible total cost incurred before the frog reaches Stone N.

Constraints

All values in input are integers.

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

1 \leq C \leq 10^{12}

1 \leq h_1 < h_2 < \cdots < h_N \leq 10^6

Input

Input is given from Standard Input in the following format:

N C
h_1 h_2 \ldots h_N

Output

Print the minimum possible total cost incurred.

Sample Input 1

5 6
1 2 3 4 5

Sample Output 1

20

If we follow the path 1 → 3 → 5, the total cost incurred would be ((3 - 1)^2 + 6) + ((5 - 3)^2 + 6) = 20.

Sample Input 2

2 1000000000000
500000 1000000

Sample Output 2

1250000000000

The answer may not fit into a 32-bit integer type.

Sample Input 3

8 5
1 3 4 5 10 11 12 13

Sample Output 3

62

If we follow the path 1 → 2 → 4 → 5 → 8, the total cost incurred would be ((3 - 1)^2 + 5) + ((5 - 3)^2 + 5) + ((10 - 5)^2 + 5) + ((13 - 10)^2 + 5) = 62.

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.


(let((a(read))(b(read))(c(read)))(write(+ b(min c(+ a b 1)))))

Score : 200 points

Problem Statement

Takahashi has A untasty cookies containing antidotes, B tasty cookies containing antidotes and C tasty cookies containing poison.

Eating a cookie containing poison results in a stomachache, and eating a cookie containing poison while having a stomachache results in a death.
As he wants to live, he cannot eat one in such a situation.
Eating a cookie containing antidotes while having a stomachache cures it, and there is no other way to cure stomachaches.

Find the maximum number of tasty cookies that Takahashi can eat.

Constraints

0 \leq A,B,C \leq 10^9

A,B and C are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the maximum number of tasty cookies that Takahashi can eat.

Sample Input 1

3 1 4

Sample Output 1

5

We can eat all tasty cookies, in the following order:

A tasty cookie containing poison

An untasty cookie containing antidotes

A tasty cookie containing poison

A tasty cookie containing antidotes

A tasty cookie containing poison

An untasty cookie containing antidotes

A tasty cookie containing poison

Sample Input 2

5 2 9

Sample Output 2

10

Sample Input 3

8 8 1

Sample Output 3

9


Score : 100 points

Problem Statement

You are given an integer N that has exactly four digits in base ten.
How many times does 2 occur in the base-ten representation of N?

Constraints

1000 \leq N \leq 9999

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

1222

Sample Output 1

3

2 occurs three times in 1222. By the way, this contest is held on December 22 (JST).

Sample Input 2

3456

Sample Output 2

0

Sample Input 3

9592

Sample Output 3

1


;; -*- coding:utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 1))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (ql:quickload :cl-debug-print))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)

;; BEGIN_INSERTED_CONTENTS
(defun make-prime-table (size)
  "Erzeugt die Primzahlentabelle 0 zu SIZE-1."
  (declare (optimize (speed 3) (safety 1)))
  (let ((dict (make-array size :element-type 'bit :initial-element 1)))
    (setf (aref dict 0) 0 (aref dict 1) 0)
    (loop for even-num from 4 below size by 2
          do (setf (aref dict even-num) 0))
    (loop for p from 3 to (ceiling (sqrt size)) by 2
          when (= 1 (aref dict p))
          do (loop for composite from (+ p p) below size by p
                   until (>= composite size)
                   do (setf (aref dict composite) 0)))
    dict))

(defun decompose-to-pow-table (num prime-table)
  (declare (optimize (speed 3) (safety 1))
           (fixnum num) ; Beachte!
           ((simple-array bit (*)) prime-table))
  (let ((factor-table (make-array (length prime-table)
                                  :element-type '(unsigned-byte 7)
                                  :initial-element 0)))
    (when (= 1 (aref prime-table 2))
      (setf (aref factor-table 2)
            (loop while (evenp num)
                  count t
                  do (setf num (ash num -1)))))
    (loop for prime from 3 to (min num (- (length prime-table) 1)) by 2
          when (= 1 (aref prime-table prime))
          do (setf (aref factor-table prime)
                   (loop while (zerop (mod num prime))
                         count t
                         do (setf num (floor num prime))))
          finally (return factor-table))))

(defmacro println (obj &optional (stream '*standard-output*))
  `(prog1 (princ ,obj ,stream) (terpri ,stream)))

;; Hauptteil

;; Sei gcd(a1, ..., aN) = 2^f(2)*3^f(3)*....
;; Da P = a1* ... *aN, gilt gcd(a1, ..., aN)^N | P und somit
;; 2^(f(2)*N)*3^(f(3)*N)*... | P
;; Wir müssen nur so ein f finden, so dass jedes f(k) maximal ist.
;; Nun sei 2^g(2)*3^g(3)*.... die Primfaktorzerlegung von P.
;; Dann muss f Folgendes erfüllen:
;; f(2)* N <= g(2)
;; f(3)* N <= g(3)...

(defun main ()
  (let* ((n (read))
         (p (read))
         (size 100001)
         (prime-dict (make-prime-table size))
         (factor-table (decompose-to-pow-table p prime-dict)))
    (if (= n 1)
        (println p)
        (loop for prime from 2 below size
              with product = 1
              when (>= (aref factor-table prime) 1)
              do (setf product
                       (* product (expt prime (floor (aref factor-table prime) n))))
              finally (println product)))))

#-swank(main)

Score : 300 points

Problem Statement

There are N integers a_1, a_2, ..., a_N not less than 1.
The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.

Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Constraints

1 \leq N \leq 10^{12}

1 \leq P \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N P

Output

Print the answer.

Sample Input 1

3 24

Sample Output 1

2

The greatest common divisor would be 2 when, for example, a_1=2, a_2=6 and a_3=2.

Sample Input 2

5 1

Sample Output 2

1

As a_i are positive integers, the only possible case is a_1 = a_2 = a_3 = a_4 = a_5 = 1.

Sample Input 3

1 111

Sample Output 3

111

Sample Input 4

4 972439611840

Sample Output 4

206


(format t "~a~%" (if (every #'evenp (loop repeat (read) collect (read))) "second" "first"))

Score : 500 points

Problem Statement

There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i.

You and Lunlun the dachshund alternately perform the following operation (starting from you):

Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors.

The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?

Constraints

1 \leq N \leq 10^5

1 \leq a_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
a_2
:
a_N

Output

If you will win, print first; if Lunlun will win, print second.

Sample Input 1

2
1
2

Sample Output 1

first

Let Color 1 be red, and Color 2 be blue. In this case, the tree bears one red apple and two blue apples.

You should eat the red apple in your first turn. Lunlun is then forced to eat one of the blue apples, and you can win by eating the other in your next turn.

Note that you are also allowed to eat two apples in your first turn, one red and one blue (not a winning move, though).

Sample Input 2

3
100000
30000
20000

Sample Output 2

second


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

;; BEGIN_INSERTED_CONTENTS
(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  "NOTE: cannot read -2^62"
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  (error "Read EOF or #\Nul."))
                                 ((= byte #.(char-code #\-))
                                  (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 brute (as)
  #>as
  (let ((n (length as)) res)
    (dotimes (pivot (+ 1 n))
      (let ((as (copy-seq as))
            (sum pivot))
        (loop for j below pivot
              do (setf (aref as j) (* -2 (aref as j))))
        (loop for j from (- pivot 1) downto 0
              unless (= j (- n 1))
              do (loop
                   (when (<= (aref as j) (aref as (+ j 1)))
                     (return))
                   (setf (aref as j) (* -2 (aref as j)))
                   (incf sum)))
        (loop for j from pivot below n
              unless (zerop j)
              do (loop
                   (when (<= (aref as (- j 1)) (aref as j))
                     (return))
                   (setf (aref as j) (* -2 (aref as j)))
                   (incf sum)))
        ;; #>as
        (push sum res)))
    (reverse res)))

(declaim (inline calc-capacity))
(defun calc-capacity (a next-a)
  (assert (<= a next-a))
  (let ((ratio (floor next-a a)))
    (ash (- (integer-length ratio) 1) -1)))

(defun main ()
  (let* ((n (read))
         (as (make-array n :element-type 'uint32)))
    (declare (uint32 n))
    (dotimes (i n)
      (setf (aref as i) (read-fixnum)))
    (labels
        ((build-cumul (as)
           (let (stack ; index . capacity
                 (cumul (make-array (+ n 1) :element-type 'uint32 :initial-element 0)))
             (push (cons (- n 1) #xffffffff) stack) ; sentinel
             (loop for i from (- n 2) downto 0
                   for next-a of-type uint32 = (aref as (+ i 1))
                   for new-value of-type uint32 = (aref cumul (+ i 1))
                   do (loop
                        (when (<= (aref as i) next-a)
                          (let ((cap (calc-capacity (aref as i) next-a)))
                            (when (> cap 0)
                              (push (cons i cap) stack)))
                          (return))
                        (setq next-a (* next-a 4))
                        (let ((node (car stack)))
                          (declare ((cons uint32 uint32) node))
                          (incf new-value (* 2 (- (car node) i)))
                          (decf (cdr node))
                          (when (zerop (cdr node))
                            (pop stack))))
                      (setf (aref cumul i) new-value))
             cumul)))
      (let ((cumul+ (build-cumul as))
            (cumul- (reverse (build-cumul (reverse as)))))
        ;; (dotimes (i (+ n 1))
        ;;   (incf (aref cumul- i) i))
        (println
         (loop for i to n
               minimize (+ (aref cumul+ i)
                           (aref cumul- i)
                           i)))))))

#-swank (main)

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

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

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

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

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

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

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

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

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "4
3 1 4 1
"
    "3
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "5
1 2 3 4 5
"
    "0
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097
"
    "7
")))

Score : 800 points

Problem Statement

There are N positive integers A_1, A_2, ..., A_N.
Takahashi can perform the following operation on these integers any number of times:

Choose 1 \leq i \leq N and multiply the value of A_i by -2.

Notice that he multiplies it by minus two.

He would like to make A_1 \leq A_2 \leq ... \leq A_N holds.
Find the minimum number of operations required. If it is impossible, print -1.

Constraints

1 \leq N \leq 200000

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 answer.

Sample Input 1

4
3 1 4 1

Sample Output 1

3

One possible solution is:

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now 3, 1, 4, -2.

Choose i=1 and multiply the value of A_1 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, -2.

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, 4.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

0

A_1 \leq A_2 \leq ... \leq A_N holds before any operation is performed.

Sample Input 3

8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097

Sample Output 3

7


(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 ()
  (declare (inline sort))
  (let* ((n (read))
         (as (make-array n :element-type 'uint31 :initial-element 0))
         (counts (make-hash-table :test #'eq))
         (res 0))
    (labels ((calc-partner (x)
               (- (sb-int:power-of-two-ceiling (+ x 1)) x)))
      (dotimes (i n)
        (setf (aref as i) (read-fixnum)))
      (setq as (sort as #'>))
      (dotimes (i n)
        (let ((a (aref as i)))
          (if (gethash a counts)
              (incf (gethash a counts))
              (setf (gethash a counts) 1))))
      (dotimes (i n)
        (let ((a (aref as i)))
          (when (> (gethash a counts) 0)
            (let ((partner (calc-partner a)))
              (when (and (gethash partner counts)
                         (> (gethash partner counts) 0))
                (if (= a partner)
                    (when (>= (gethash partner counts) 2)
                      (incf res)
                      (decf (gethash partner counts) 2))
                    (progn
                      (incf res)
                      (decf (gethash a counts))
                      (decf (gethash partner counts)))))))))
      (println 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 "3
1 2 3
"
    "1
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "5
3 11 14 5 13
"
    "2
")))

Score : 600 points

Problem Statement

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i.
He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2.
Note that a ball cannot belong to multiple pairs.
Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

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

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2


(in-package :cl-user)
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx opt
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (set-dispatch-macro-character #\# #\> #'cl-debug-print:debug-print-reader)

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

(defconstant +mod+ 1000000007)

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

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

;; BEGIN_INSERTED_CONTENTS
;; BEGIN_USE_PACKAGE
(in-package :cl-user)

;;;
;;; Body
;;;

(defun main ()
  (let* ((n (read))
         (ps (sort (loop repeat n collect (read)) #'>)))
    (println (+ (floor (car ps) 2)
                (reduce #'+ (cdr ps))))))

#-swank (main)

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

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

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

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

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

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

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

#-swank
(eval-when (:compile-toplevel)
  (when (and (boundp 'sb-c::*compiler-warning-count*)
             (> sb-c::*compiler-warning-count* 0))
    (sb-ext:quit :unix-status 1)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (equal "15950
"
          (run "3
4980
7980
6980
" nil)))
  (it.bese.fiveam:is
   (equal "15120
"
          (run "4
4320
4320
4320
4320
" nil))))

Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.


#-swank
(unless (member :child-sbcl *features*)
  "I refered from https://competitive12.blogspot.com/2020/03/common-lisp.html thank you!"
  (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))))
#-swank
(unless (member :child-sbcl *features*)
  "I refered from https://competitive12.blogspot.com/2020/03/common-lisp.html thank you!"
  (quit
   :unix-status
   (process-exit-code
    (run-program *runtime-pathname*
                 `("--control-stack-size" "256MB"
                   "--noinform" "--disable-ldb" "--lose-on-corruption" "--end-runtime-options"
                   "--eval" "(push :child-sbcl *features*)"
                   "--script" ,(namestring *load-pathname*))
                 :output t :error t :input t))))
(defmacro dp (name args memo-size &body body)
  (let ((memo (gensym)))
    `(let ((,memo (make-array ,memo-size :initial-element nil)))
       (defun ,name ,args
         (or (aref ,memo ,@args)
             (setf (aref ,memo ,@args)
                   ,@body))))))

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

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

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

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

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

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

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

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

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

(defun range-0-n (n &optional (step 1))
  (loop for i from 0 below n by step collect i))

(defun range-1-n (n &optional (step 1))
  (loop for i from 1 below n by step collect i))

(defun range-a-b (a b &optional (step 1))
  (loop for i from a below b by step collect i))

(defun map-0-n (function n &optional (step 1))
  (mapcar function (range-0-n n step)))

(defun map-1-n (function n &optional (step 1))
  (mapcar function (range-1-n n step)))

(defun map-a-b (function a b &optional (step 1))
  (mapcar function (range-a-b a b step)))

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

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

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

(defun nearby (&rest args)
  (let ((current (subseq args 0 (ash (length args) -1)))
        (validator (subseq args (ash (length args) -1)))
        (res nil))
    (labels ((check ()
               (every (lambda (x y) (and (<= 0 x) (< x y)))
                      current validator))
             (f (lst)
               (unless lst (return-from f))
               (let ((x (car lst)))
                 (setf (car lst) (1+ x))
                 (when (check) (push (copy-list current) res))
                 (setf (car lst) (1- x))
                 (when (check) (push (copy-list current) res))
                 (setf (car lst) x))
               (f (cdr lst))))
      (f current)
      res)))

(defun main (line)
  (- (reduce #'+ line) (floor (reduce #'max line) 2)))

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


(let* ((n (read))
    (k (read))
    (h (make-array n))
    (ans 0))
    (loop for i below n do
        (setf (aref h i) (read))
    )
    (setf h (sort h '>))
    (loop for i to (- n k) do
        (if (zerop i)
            (setq ans (- (aref h i) (aref h (+ i (- k 1)))))
            (if (> ans (- (aref h i) (aref h (+ i (- k 1)))))
                (setq ans (- (aref h i) (aref h (+ i (- k 1)))))
            )
        )
    )
    (format t "~A~%" ans)
)

Score : 300 points

Problem Statement

In some other world, today is Christmas Eve.

There are N trees planted in Mr. Takaha's garden. The height of the i-th tree (1 \leq i \leq N) is h_i meters.

He decides to choose K trees from these trees and decorate them with electric lights. To make the scenery more beautiful, the heights of the decorated trees should be as close to each other as possible.

More specifically, let the height of the tallest decorated tree be h_{max} meters, and the height of the shortest decorated tree be h_{min} meters. The smaller the value h_{max} - h_{min} is, the better. What is the minimum possible value of h_{max} - h_{min}?

Constraints

2 \leq K < N \leq 10^5

1 \leq h_i \leq 10^9

h_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
h_1
h_2
:
h_N

Output

Print the minimum possible value of h_{max} - h_{min}.

Sample Input 1

5 3
10
15
11
14
12

Sample Output 1

2

If we decorate the first, third and fifth trees, h_{max} = 12, h_{min} = 10 so h_{max} - h_{min} = 2. This is optimal.

Sample Input 2

5 3
5
7
5
7
7

Sample Output 2

0

If we decorate the second, fourth and fifth trees, h_{max} = 7, h_{min} = 7 so h_{max} - h_{min} = 0. This is optimal.

There are not too many trees in these sample inputs, but note that there can be at most one hundred thousand trees (we just can't put a sample with a hundred thousand lines here).


(defun solve(n k h)
    (declare (inline sort))
  (declare (inline sb-impl::stable-sort-list))
    (let ((_h (make-array (length h) :initial-contents (sort h #'<=)))
	  (ans (expt 10 9)))
      (dotimes (i (1+ (- n k)) ans)
	(setf ans (min ans (- (aref _h (+ i (1- k)))
			      (aref _h i)))))))

(defun main ()
  (let* ((n (read))
	 (k (read))
	 (h (make-list n)))
    (dotimes (i n)
      (setf (nth i h) (read)))
    (princ (solve n k h))
    (fresh-line)))

(main)


(in-package :cl-user)
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx opt
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (set-dispatch-macro-character #\# #\> #'cl-debug-print:debug-print-reader)

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

(defconstant +mod+ 1000000007)

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

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

;; BEGIN_INSERTED_CONTENTS
;; BEGIN_USE_PACKAGE
(in-package :cl-user)

;;;
;;; Body
;;;

(defun main ()
  (let* ((n (read))
         (x (read)))
    (labels ((calc-length (level)
               (let ((res 1))
                 (dotimes (_ level res)
                   (setq res (+ 3 (* res 2))))))
             (calc-patty (level)
               (let ((res 1))
                 (dotimes (_ level res)
                   (setq res (+ 1 (* res 2))))))
             (recur (level len)
               (dbg level len)
               (cond ((>= (+ len 1) x) 0)
                     ((>= (+ len 1 (calc-length (- level 1))) x)
                      (recur (- level 1) (+ len 1)))
                     (t (+ (calc-patty (- level 1))
                           1
                           (recur (- level 1) (+ len (calc-length (- level 1)) 2)))))))
      (println
       (recur n 0)))))

#-swank (main)

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

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

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

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

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

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

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

#-swank
(eval-when (:compile-toplevel)
  (when (and (boundp 'sb-c::*compiler-warning-count*)
             (> sb-c::*compiler-warning-count* 0))
    (sb-ext:quit :unix-status 1)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (equal "4
"
          (run "2 7
" nil)))
  (it.bese.fiveam:is
   (equal "0
"
          (run "1 1
" nil)))
  (it.bese.fiveam:is
   (equal "2160549382716056
"
          (run "50 4321098765432109
" nil))))

Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

A level-0 burger is a patty.

A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

1 \leq N \leq 50

1 \leq X \leq ( the total number of layers in a level-N burger )

N and X are integers.

Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.


(defun solve (x)
  (if (or (= x 7) (= x 5) (= x 3))
    "YES"
    "NO"))
(format t "~A~%" (solve (read)))

Score : 100 points

Problem Statement

Shichi-Go-San (literally "Seven-Five-Three") is a traditional event in a certain country to celebrate the growth of seven-, five- and three-year-old children.

Takahashi is now X years old. Will his growth be celebrated in Shichi-Go-San this time?

Constraints

1 ≤ X ≤ 9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

If Takahashi's growth will be celebrated, print YES; if it will not, print NO.

Sample Input 1

5

Sample Output 1

YES

The growth of a five-year-old child will be celebrated.

Sample Input 2

6

Sample Output 2

NO

See you next year.


(in-package :cl-user)
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx opt
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (set-dispatch-macro-character #\# #\> #'cl-debug-print:debug-print-reader)

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

(defconstant +mod+ 1000000007)

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

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

;; BEGIN_INSERTED_CONTENTS
;; BEGIN_USE_PACKAGE
(in-package :cl-user)

;;;
;;; Body
;;;

(defun main ()
  (let* ((s (read-line))
         (n (length s)))
    (println
     (loop for i from 0 to (- n 3)
           for x = (read-from-string s t nil :start i :end (+ i 3))
           minimize (abs (- x 753))))))

#-swank (main)

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

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

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

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

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

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

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

#-swank
(eval-when (:compile-toplevel)
  (when (and (boundp 'sb-c::*compiler-warning-count*)
             (> sb-c::*compiler-warning-count* 0))
    (sb-ext:quit :unix-status 1)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (5am:is
   (equal "34
"
          (run "1234567876
" nil)))
  (5am:is
   (equal "0
"
          (run "35753
" nil)))
  (5am:is
   (equal "642
"
          (run "1111111111
" nil))))

Score : 200 points

Problem Statement

There is a string S consisting of digits 1, 2, ..., 9.
Lunlun, the Dachshund, will take out three consecutive digits from S, treat them as a single integer X and bring it to her master. (She cannot rearrange the digits.)

The master's favorite number is 753. The closer to this number, the better.
What is the minimum possible (absolute) difference between X and 753?

Constraints

S is a string of length between 4 and 10 (inclusive).

Each character in S is 1, 2, ..., or 9.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum possible difference between X and 753.

Sample Input 1

1234567876

Sample Output 1

34

Taking out the seventh to ninth characters results in X = 787, and the difference between this and 753 is 787 - 753 = 34. The difference cannot be made smaller, no matter where X is taken from.

Note that the digits cannot be rearranged. For example, taking out 567 and rearranging it to 765 is not allowed.

We cannot take out three digits that are not consecutive from S, either. For example, taking out the seventh digit 7, the ninth digit 7 and the tenth digit 6 to obtain 776 is not allowed.

Sample Input 2

35753

Sample Output 2

0

If 753 itself can be taken out, the answer is 0.

Sample Input 3

1111111111

Sample Output 3

642

No matter where X is taken from, X = 111, with the difference 753 - 111 = 642.


#+swank (declaim (optimize (speed 0) (safety 3) (debug 3)))
#-swank (declaim (optimize (speed 3) (safety 0) (debug 0)))

(defconstant +mod+ 1000000007)

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

(defmethod fast-sort ((sequence list) &key (test #'<))
  (declare (inline sort)
           (inline sb-impl::stable-sort-list))
  (sort sequence (lambda (x y)
                   (funcall test x y))))

(defmethod fast-sort ((sequence array) &key (test #'<))
  (declare (inline sort))
  (sort sequence (lambda (x y)
                   (funcall test x y))))

(defmacro read-numbers-to-list (size)
  `(progn
     (when (not (integerp ,size))
       (error "Size must be integer."))
     (when (< ,size 0)
       (error "Size must be plus or zero."))
     (loop repeat ,size collect (read))))

(defmacro read-numbers-to-array (size)
  `(progn
     (when (not (integerp ,size))
       (error "Size must be integer."))
     (when (< ,size 0)
       (error "Size must be plus or zero."))
     (make-array ,size :initial-contents (read-numbers-to-list ,size))))

(defmacro read-numbers-to-board (row-size column-size)
  (let ((board (gensym))
        (r (gensym))
        (c (gensym)))
    `(let ((,board (make-array '(,row-size ,column-size))))
       (dotimes (,r ,row-size)
         (dotimes (,c ,column-size)
           (setf (aref ,board ,r ,c) (read))))
       ,board)))

(defmethod make-cumlative-sum ((sequence list))
  (labels ((inner (sequence &optional (acc '(0)))
             (if (null sequence)
                 (reverse acc)
                 (inner (rest sequence) (cons (+ (first sequence)
                                                 (first acc))
                                              acc)))))
    (inner sequence)))

(defmethod make-cumlative-sum ((sequence array))
  (declare (type (simple-array fixnum) sequence))
  (the array
       (let* ((n (length sequence))
              (acc (make-array (1+ n) :element-type 'integer :initial-element 0)))
         (loop for i below n do
              (setf (aref acc (1+ i)) (+ (aref sequence i)
                                         (aref acc i)))
            finally
              (return acc)))))

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

(defun unwrap (list)
  (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)))))

;;; Write code here

(defun int->lst (k &optional (acc nil))
  (if (zerop k)
      acc
      (int->lst
       (truncate k 10)
       (cons (rem k 10) acc))))

(defun lst->int (xs &optional (acc 0))
  (If (null xs)
      acc
      (let ((tmp (pop xs)))
        (lst->int xs (+ (* acc 10) tmp)))))

(defun judge (k)
  (every
   (lambda (x) (find x (int->lst k)))
   '(3 5 7)))

(defun solve (n)
  (labels ((inner (k)
             (if (> k n)
                 0
                 (+
                  (if (judge k) 1 0)
                  (reduce #'+
                          (mapcar (lambda (c)
                                    (inner (+ (* k 10)
                                              c)))
                                  '(3 5 7)))))))
    (inner 0)))

(defun main ()
  (let ((n (read)))
    (format t "~a~%" (solve n))))

(main)

Score : 300 points

Problem Statement

You are given an integer N. Among the integers between 1 and N (inclusive), how many Shichi-Go-San numbers (literally "Seven-Five-Three numbers") are there?

Here, a Shichi-Go-San number is a positive integer that satisfies the following condition:

When the number is written in base ten, each of the digits 7, 5 and 3 appears at least once, and the other digits never appear.

Constraints

1 \leq N < 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go-San numbers between 1 and N (inclusive).

Sample Input 1

575

Sample Output 1

4

There are four Shichi-Go-San numbers not greater than 575: 357, 375, 537 and 573.

Sample Input 2

3600

Sample Output 2

13

There are 13 Shichi-Go-San numbers not greater than 3600: the above four numbers, 735, 753, 3357, 3375, 3537, 3557, 3573, 3575 and 3577.

Sample Input 3

999999999

Sample Output 3

26484


(let ((n (read))
      (k (read))
      (a (make-array 1 :element-type 'integer
                       :initial-element 0
                       :fill-pointer 1
                       :adjustable t))
      (b (make-array 0 :element-type 'integer
                       :adjustable t
                       :fill-pointer 0))
      (ans 0))
  (loop for x from 1 upto n do
        (vector-push-extend (+ (aref a (1- x)) (read)) a))
  (loop for x from 0 upto (1- n) do
        (loop for y from (1+ x) upto n do
              (vector-push-extend (- (aref a y) (aref a x)) b)))
  (loop for i from 40 downto 0 do
        (let ((count 0))
          (loop for x across b do
                (if (= (+ ans (expt 2 i)) (logand x (+ ans (expt 2 i)))) (incf count)))
          (if (>= count k) (incf ans (expt 2 i)))))
  (princ ans))

Score : 400 points

Problem Statement

One day, Niwango-kun, an employee of Dwango Co., Ltd., found an integer sequence (a_1, ..., a_N) of length N.
He is interested in properties of the sequence a.

For a nonempty contiguous subsequence a_l, ..., a_r (1 \leq l \leq r \leq N) of the sequence a, its beauty is defined as a_l + ... + a_r. Niwango-kun wants to know the maximum possible value of the bitwise AND of the beauties of K nonempty contiguous subsequences among all N(N+1)/2 nonempty contiguous subsequences. (Subsequences may share elements.)

Find the maximum possible value for him.

Constraints

2 \leq N \leq 1000

1 \leq a_i \leq 10^9

1 \leq K \leq N(N+1)/2

All numbers given in input are integers

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 ... a_N

Output

Print the answer.

Sample Input 1

4 2
2 5 2 5

Sample Output 1

12

There are 10 nonempty contiguous subsequences of a. Let us enumerate them:

contiguous subsequences starting from the first element: \{2\}, \{2, 5\}, \{2, 5, 2\}, \{2, 5, 2, 5\}

contiguous subsequences starting from the second element: \{5\}, \{5, 2\}, \{5, 2, 5\}

contiguous subsequences starting from the third element: \{2\}, \{2, 5\}

contiguous subsequences starting from the fourth element: \{5\}

(Note that even if the elements of subsequences are equal, subsequences that have different starting indices are considered to be different.)

The maximum possible bitwise AND of the beauties of two different contiguous subsequences is 12.
This can be achieved by choosing \{5, 2, 5\} (with beauty 12) and \{2, 5, 2, 5\} (with beauty 14).

Sample Input 2

8 4
9 1 8 2 7 5 6 4

Sample Output 2

32


;; -*- 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 nil t))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
(declaim (inline read-line-into))
(defun read-line-into (buffer-string &key (in *standard-input*) (term-char #\Space))
  "Receives ASCII inputs and returns multiple values: the string and the end
position.

This function calls READ-BYTE to read characters though it calls READ-CHAR
instead on SLIME because SLIME's IO is not bivalent."
  (declare (inline read-byte)
           #-swank (sb-kernel:ansi-stream in))
  (loop for c of-type base-char =
           #-swank (code-char (read-byte in nil #.(char-code #\Newline)))
           #+swank (read-char in nil #\Newline)
        for idx from 0
        until (char= c #\Newline)
        do (setf (char buffer-string idx) c)
        finally (when (< idx (length buffer-string))
                  (setf (char buffer-string idx) term-char))
                (return (values buffer-string idx))))

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

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

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

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun main ()
  (declare #.OPT)
  (let* ((n (read))
         (s (let ((s (make-string n :element-type 'base-char)))
              (read-line-into s)))
         (q (read))
         (ks (make-array q :element-type 'uint32))
         (cumul-d (make-array (+ n 1) :element-type 'uint32))
         (cumul-m (make-array (+ n 1) :element-type 'uint32))
         (cumul-c (make-array (+ n 1) :element-type 'uint32))
         (cumul-dm (make-array (+ n 1) :element-type 'uint62)))
    (declare (uint31 n q)
             (simple-base-string s))
    (dotimes (i q)
      (setf (aref ks i) (read)))
    (dotimes (i n)
      (setf (aref cumul-d (+ i 1))
            (+ (aref cumul-d i)
               (if (char= #\D (aref s i)) 1 0)))
      (setf (aref cumul-m (+ i 1))
            (+ (aref cumul-m i)
               (if (char= #\M (aref s i)) 1 0)))
      (setf (aref cumul-c (+ i 1))
            (+ (aref cumul-c i)
               (if (char= #\C (aref s i)) 1 0))))
    (dotimes (i n)
      (setf (aref cumul-dm (+ i 1))
            (if (char= #\M (aref s i))
                (+ (aref cumul-dm i) (aref cumul-d i))
                (aref cumul-dm i))))
    (let ((total-dmc 0))
      (declare (uint62 total-dmc))
      (dotimes (i n)
        (when (char= #\M (aref s i))
          (incf total-dmc (* (aref cumul-d i)
                             (- (aref cumul-c n) (aref cumul-c i))))))
      (labels ((calc (k)
                 (let ((res 0))
                   (declare (uint62 res))
                   (dotimes (z n)
                     (when (char= #\C (aref s z))
                       (incf res (aref cumul-dm (max 0 (+ 1 (- z k)))))
                       (incf res (* (aref cumul-d (max 0 (+ 1 (- z k))))
                                    (- (aref cumul-m z)
                                       (aref cumul-m (max 0 (+ 1 (- z k)))))))))
                   res)))
        (sb-int:dovector (k ks)
          (println (- total-dmc (calc k))))))))

#-swank (main)

Score : 600 points

Problem Statement

In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.

The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.

Given a string S of length N and an integer k (k \geq 3),
he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:

0 \leq a < b < c \leq N - 1

S[a] = D

S[b] = M

S[c] = C

c-a < k

Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.

For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).

Constraints

3 \leq N \leq 10^6

S consists of uppercase English letters

1 \leq Q \leq 75

3 \leq k_i \leq N

All numbers given in input are integers

Input

Input is given from Standard Input in the following format:

N
S
Q
k_{0} k_{1} ... k_{Q-1}

Output

Print Q lines.
The i-th line should contain the k_i-DMC number of the string S.

Sample Input 1

18
DWANGOMEDIACLUSTER
1
18

Sample Output 1

1

(a,b,c) = (0, 6, 11) satisfies the conditions.

Strangely, Dwango Media Cluster does not have so much DMC-ness by his definition.

Sample Input 2

18
DDDDDDMMMMMCCCCCCC
1
18

Sample Output 2

210

The number of triples can be calculated as 6\times 5\times 7.

Sample Input 3

54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40

Sample Output 3

0
1
2

(a, b, c) = (0, 23, 36), (8, 23, 36) satisfy the conditions except the last one, namely, c-a < k_i.

By the way, DWANGO is an acronym for "Dial-up Wide Area Network Gaming Operation".

Sample Output 4

30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20

Sample Output 4

10
52
110
140


(in-package :cl-user)
(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx opt
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (set-dispatch-macro-character #\# #\> #'cl-debug-print:debug-print-reader)

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

(defconstant +mod+ 1000000007)

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

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

;; BEGIN_INSERTED_CONTENTS
;; BEGIN_USE_PACKAGE
(in-package :cl-user)

;;;
;;; Body
;;;

(defun main ()
  (let* ((x (read))
         (y (read)))
    (println (+ x (floor y 2)))))

#-swank (main)

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

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

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

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

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

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

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

#-swank
(eval-when (:compile-toplevel)
  (when (and (boundp 'sb-c::*compiler-warning-count*)
             (> sb-c::*compiler-warning-count* 0))
    (sb-ext:quit :unix-status 1)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (5am:is
   (equal "110
"
          (run "81 58
" nil)))
  (5am:is
   (equal "31
"
          (run "4 54
" nil))))

Score: 100 points

Problem Statement

There is a train going from Station A to Station B that costs X yen (the currency of Japan).

Also, there is a bus going from Station B to Station C that costs Y yen.

Joisino got a special ticket. With this ticket, she can take the bus for half the fare if she travels from Station A to Station B by train and then travels from Station B to Station C by bus.

How much does it cost to travel from Station A to Station C if she uses this ticket?

Constraints

1 \leq X,Y \leq 100

Y is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

If it costs x yen to travel from Station A to Station C, print x.

Sample Input 1

81 58

Sample Output 1

110

The train fare is 81 yen.

The train fare is 58 ⁄ 2=29 yen with the 50% discount.

Thus, it costs 110 yen to travel from Station A to Station C.

Sample Input 2

4 54

Sample Output 2

31


(let ((x (read))
      (y (read)))
     (princ (+ x (floor y 2))
            ))


(defun f (X Y)
  (format t "~a" (+ X (/ Y 2))))

(f ((read) (read)))


(let* ((n (read))
       (tm (read))
       (a (read))
       (h (loop :repeat n :collect(read))))
  (map-into h (lambda (w)
            (abs (- a (- tm (* w 0.006))))) h)
  (princ (1+ (position (reduce #'min h) h))))

Score: 200 points

Problem Statement

A country decides to build a palace.

In this country, the average temperature of a point at an elevation of x meters is T-x \times 0.006 degrees Celsius.

There are N places proposed for the place. The elevation of Place i is H_i meters.

Among them, Princess Joisino orders you to select the place whose average temperature is the closest to A degrees Celsius, and build the palace there.

Print the index of the place where the palace should be built.

It is guaranteed that the solution is unique.

Constraints

1 \leq N \leq 1000

0 \leq T \leq 50

-60 \leq A \leq T

0 \leq H_i \leq 10^5

All values in input are integers.

The solution is unique.

Input

Input is given from Standard Input in the following format:

N
T A
H_1 H_2 ... H_N

Output

Print the index of the place where the palace should be built.

Sample Input 1

2
12 5
1000 2000

Sample Output 1

1

The average temperature of Place 1 is 12-1000 \times 0.006=6 degrees Celsius.

The average temperature of Place 2 is 12-2000 \times 0.006=0 degrees Celsius.

Thus, the palace should be built at Place 1.

Sample Input 2

3
21 -11
81234 94124 52141

Sample Output 2

3


;; -*- coding: utf-8 -*-
(eval-when (:compile-toplevel :load-toplevel :execute)
  (defparameter OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0)))
  #+swank (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
(defmacro with-output-buffer (&body body)
  "Buffers all outputs to *STANDARD-OUTPUT* and flushes them to
*STANDARD-OUTPUT* at the end. Note that only BASE-CHAR is allowed."
  (let ((out (gensym)))
    `(let ((,out (make-string-output-stream :element-type 'base-char)))
       (let ((*standard-output* ,out))
         ,@body)
       (write-string (get-output-stream-string ,out)))))

(declaim (ftype (function * (values fixnum &optional)) read-fixnum))
(defun read-fixnum (&optional (in *standard-input*))
  (declare #.OPT)
  (macrolet ((%read-byte ()
               `(the (unsigned-byte 8)
                     #+swank (char-code (read-char in nil #\Nul))
                     #-swank (sb-impl::ansi-stream-read-byte in nil #.(char-code #\Nul) nil))))
    (let* ((minus nil)
           (result (loop (let ((byte (%read-byte)))
                           (cond ((<= 48 byte 57)
                                  (return (- byte 48)))
                                 ((zerop byte) ; #\Nul
                                  ;; (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 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)

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

;; Hauptteil

(defstruct city
  (year 0 :type uint31)
  (prefecture 0 :type uint31)
  (version 0 :type uint31))

(defun main ()
  (declare #.OPT
           (inline sort))
  (let* ((n (read))
         (m (read))
         (p-table (make-array (1+ n) :element-type 'list :initial-element nil))
         (city-table (make-array m :element-type 'list :initial-element nil)))
    (declare (uint32 n m))
    (dotimes (city-idx m)
      (let* ((prefecture (read-fixnum))
             (year (read-fixnum))
             (node (make-city :prefecture prefecture :year year)))
        (push node (aref p-table prefecture))
        (setf (aref city-table city-idx) node)))
    (loop for p-idx from 1 to n
          do (setf (aref p-table p-idx)
                   (sort (the list (aref p-table p-idx)) #'< :key #'city-year))
             (loop for node in (aref p-table p-idx)
                   for version from 1
                   do (setf (city-version node) version)))
    (with-output-buffer
      (dotimes (city-idx m)
        (format t "~6,'0D~6,'0D~%"
                (city-prefecture (aref city-table city-idx))
                (city-version (aref city-table city-idx)))))))

#-swank(main)

Score: 300 points

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Y_i and belongs to Prefecture P_i.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is P_i, and the last six digits of the ID number is x.

Here, if P_i or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq P_i \leq N

1 \leq Y_i \leq 10^9

Y_i are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
P_1 Y_1
:
P_M Y_M

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, ...).

Sample Input 1

2 3
1 32
2 63
1 12

Sample Output 1

000001000002
000002000001
000001000001

As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.

As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.

As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

Sample Input 2

2 3
2 55
2 77
2 99

Sample Output 2

000002000001
000002000002
000002000003


(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
(defmacro split-and-bind (arg-lst string &body body)
  (let ((pos1 (gensym "POS"))
	(pos2 (gensym "POS"))
	(str (gensym "STR")))
    (labels ((expand (arg-lst &optional (init-pos1 t))
	       (if (null arg-lst)
		   body
		   `((let* ((,pos1 ,(if init-pos1 0 `(1+ ,pos2)))
			    (,pos2 (position #\space ,str :start ,pos1 :test #'char=))
			    (,(car arg-lst) (parse-integer ,str :start ,pos1 :end ,pos2)))
		       ,@(expand (cdr arg-lst) nil))))))
      `(let ((,str ,string))
         (declare (string ,str))
	 ,@(expand arg-lst)))))

(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 4 7 8 15 16 31 32 62 63 64)

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

;; Hauptteil

(defstruct city
  (year 0 :type uint31)
  (prefecture 0 :type uint31)
  (version 0 :type uint31))

(defun main ()
  (let* ((n (read))
         (m (read))
         (p-table (make-array (list (1+ n)) :element-type 'list :initial-element nil))
         (city-table (make-array (list m) :element-type 'list :initial-element nil)))
    ;; node := (prefecture year version)
    (dotimes (city-idx m)
      (split-and-bind (prefecture year) (read-line)
        (let ((node (make-city :prefecture prefecture :year year)))
          (push node (aref p-table prefecture))
          (setf (aref city-table city-idx) node))))
    (loop for p-idx from 1 to n
          do (setf (aref p-table p-idx)
                   (sort (aref p-table p-idx) #'< :key #'city-year))
             (loop for node in (aref p-table p-idx)
                   for version from 1
                   do (setf (city-version node) version)))
    (dotimes (city-idx m)
      (format t "~6,'0D~6,'0D~%"
              (city-prefecture (aref city-table city-idx))
              (city-version (aref city-table city-idx))))))

#-swank(main)


(let* ((a nil)
       (b nil)
       (n (read))
       (m (read))
       (citylank (loop :repeat (1+ n) :collect '())))
  (loop :repeat m :do(push (read) a) :do(push (read) b))
  (setf a (reverse a))
  (setf b (reverse b))
  (mapcar (lambda (p q)
            (push q (nth p citylank))) a b)
  (mapcar (lambda (li) (sort li #'<)) citylank)
  (mapcar (lambda (c d)
            (m-print c (1+ (position d (nth c citylank))))) a b))


(let ((lst (concatenate 'list (read-line))))
  (if (= 2 (length lst))
      (format t "~A" (concatenate 'string lst))
      (format t "~A" (concatenate 'string (reverse lst)))))

Score : 100 points

Problem Statement

You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.

Constraints

The length of S is 2 or 3.

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If the length of S is 2, print S as is; if the length is 3, print S after reversing it.

Sample Input 1

abc

Sample Output 1

cba

As the length of S is 3, we print it after reversing it.

Sample Input 2

ac

Sample Output 2

ac

As the length of S is 2, we print it as is.