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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

# -*- coding: utf-8 -*-
import math
import sys
buff_readline = sys.stdin.buffer.readline
readline = sys.stdin.readline

INF = 2**60


def gen(a, b, num):
    import random
    result = set()
    while 1:
        while 1:
            v = random.randint(a, b)//2*2+1
            if v not in result:
                break
        for x in range(3, int(math.sqrt(v))+1, 2):
            if v % x == 0:
                break
        else:
            result.add(v)
            if len(result) == num:
                break
    return result


class RH():
    def __init__(self, s, base, mod):
        self.base = base
        self.mod = mod
        self.rev = pow(base, mod-2, mod)

        l = len(s)
        self.h = h = [0]*(l+1)
        tmp = 0
        for i in range(l):
            num = ord(s[i])
            tmp = (tmp*base + num) % mod
            h[i+1] = tmp

        self.pw = pw = [1]*(len(s)+1)
        v = 1
        for i in range(l):
            pw[i+1] = v = v * base % mod

    def calc(self, l, r):
        return (self.h[r] - self.h[l] * self.pw[r-l]) % self.mod


class RRH():
    def __init__(self, s, rand=False, num=5):
        if rand:
            bases = gen(2, 10**3, num)
        else:
            bases = {293, 389, 167}

        MOD = 10**9+7
        self.rhs = [RH(s, b, MOD) for b in bases]

    def calc(self, l, r):
        return [rh.calc(l, r) for rh in self.rhs]


def slv(S, T):
    lt = len(T)
    ls = len(S)
    SS = S * (-(-lt // ls) * 2)
    srh = RRH(SS)
    th = RRH(T).calc(0, len(T))

    ok = set()
    for i in range(ls):
        if srh.calc(i, i+lt) == th:
            ok.add(i)

    ans = 0
    for s in ok:
        done = set()
        t = 1
        while True:
            if s in done:
                return -1
            elif (s + lt) % ls in ok:
                done.add(s)
                s = (s + lt) % ls
                t += 1
            else:
                break
        ans = max(ans, t)
    return ans


def main():
    S = input()
    T = input()
    print(slv(S, T))


if __name__ == '__main__':
    main()