Problem:
Score : 600 points

Problem Statement

Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.

There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.

Notes

A string a is a substring of another string b if and only if there exists an integer x (0 \leq x \leq |b| - |a|) such that, for any y (1 \leq y \leq |a|), a_y = b_{x+y} holds.

We assume that the concatenation of zero copies of any string is the empty string. From the definition above, the empty string is a substring of any string. Thus, for any two strings s and t, i = 0 satisfies the condition in the problem statement.

Constraints

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

1 \leq |t| \leq 5 \times 10^5

s and t consist of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print -1.

Sample Input 1

abcabab
ab

Sample Output 1

3

The concatenation of three copies of t, ababab, is a substring of the concatenation of two copies of s, abcabababcabab, so i = 3 satisfies the condition.

On the other hand, the concatenation of four copies of t, abababab, is not a substring of the concatenation of any number of copies of s, so i = 4 does not satisfy the condition.

Similarly, any integer greater than 4 does not satisfy the condition, either. Thus, the number of non-negative integers i satisfying the condition is finite, and the maximum value of such i is 3.

Sample Input 2

aa
aaaaaaa

Sample Output 2

-1

For any non-negative integer i, the concatenation of i copies of t is a substring of the concatenation of 4i copies of s. Thus, there are infinitely many non-negative integers i that satisfy the condition.

Sample Input 3

aba
baaab

Sample Output 3

0

As stated in Notes, i = 0 always satisfies the condition.

Solution in Common Lisp (SBCL 1.1.14) (judge verdict: Wrong Answer)

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

;; BEGIN_INSERTED_CONTENTS
;;;
;;; Rolling hash (32-bit)
;;; Use 62-bit version instead. I leave it just for my reference.
;;;

(defstruct (rhash (:constructor %make-rhash (modulus base cumul powers)))
  (modulus 4294967291 :type (unsigned-byte 32))
  (base 2095716802 :type (unsigned-byte 32))
  (cumul nil :type (simple-array (unsigned-byte 32) (*)))
  (powers nil :type (simple-array (unsigned-byte 32) (*))))

(declaim (inline make-rhash))
(defun make-rhash (vector modulus &key (key #'char-code) base)
  "Returns the table of rolling-hash of VECTOR modulo MODULUS. KEY is applied to
  each element of VECTOR prior to computing the hash value.

MODULUS := unsigned 32-bit prime number
BASE := 1 | 2 | ... | MODULUS - 1
KEY := function returning FIXNUM"
  (declare (vector vector)
           ((unsigned-byte 32) modulus)
           ((or null (unsigned-byte 32)) base)
           (function key))
  (assert (sb-int:positive-primep modulus))
  (let* ((base (or base (+ 1 (random (- modulus 1)))))
         (size (length vector))
         (cumul (make-array (+ 1 size) :element-type '(unsigned-byte 32)))
         (powers (make-array (+ 1 size) :element-type '(unsigned-byte 32))))
    (assert (<= 1 base (- modulus 1)))
    (setf (aref powers 0) 1)
    (dotimes (i size)
      (setf (aref powers (+ i 1))
            (mod (* (aref powers i) base) modulus))
      (let ((sum (+ (mod (* (aref cumul i) base) modulus)
                    (mod (the fixnum (funcall key (aref vector i))) modulus))))
        (setf (aref cumul (+ i 1))
              (if (> sum modulus)
                  (- sum modulus)
                  sum))))
    (%make-rhash modulus base cumul powers)))

(declaim (inline rhash-query)
         (ftype (function * (values (unsigned-byte 32) &optional)) rhash-query))
(defun rhash-query (rhash l r)
  "Returns the hash value of the interval [L, R)."
  (declare ((integer 0 #.most-positive-fixnum) l r))
  (assert (<= l r))
  (let ((cumul (rhash-cumul rhash))
        (powers (rhash-powers rhash))
        (modulus (rhash-modulus rhash)))
    (let ((res (+ (aref cumul r)
                  (- modulus (mod (* (aref cumul l) (aref powers (- r l))) modulus)))))
      (if (> res modulus)
          (- res modulus)
          res))))

(declaim (ftype (function * (values (unsigned-byte 32) &optional)) rhash-vector-hash)
         (inline rhash-vector-hash))
(defun rhash-vector-hash (rhash vector &key (key #'char-code))
  "Returns the hash code of VECTOR w.r.t. the moduli and bases of RHASH."
  (declare (vector vector))
  (let* ((mod (rhash-modulus rhash))
         (base (rhash-base rhash))
         (size (length vector))
         (result 0))
    (declare ((unsigned-byte 32) result))
    (dotimes (i size)
      ;; (2^32-1) * (2^32-1) + (2^32-1) < 2^64
      (setq result (mod (+ (* base result)
                           (the (unsigned-byte 32)
                                (mod (the fixnum (funcall key (aref vector i))) mod)))
                        mod)))
    result))

(declaim (inline rhash-get-lcp))
(defun rhash-get-lcp (rhash1 start1 rhash2 start2)
  (declare (optimize (speed 3))
           ((integer 0 #.most-positive-fixnum) start1 start2))
  (let ((max-length (min (- (length (rhash-cumul rhash1)) start1 1)
                         (- (length (rhash-cumul rhash2)) start2 1))))
    (declare (optimize (safety 0)))
    (if (= (rhash-query rhash1 start1 (+ start1 max-length))
           (rhash-query rhash2 start2 (+ start2 max-length)))
        max-length
        (labels ((bisect (ok ng)
                   (declare ((integer 0 #.most-positive-fixnum) ok ng))
                   (if (<= (- ng ok) 1)
                       ok
                       (let ((mid (ash (+ ng ok) -1)))
                         (if (= (rhash-query rhash1 start1 (+ start1 mid))
                                (rhash-query rhash2 start2 (+ start2 mid)))
                             (bisect mid ng)
                             (bisect ok mid))))))
          (bisect 0 max-length)))))