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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys

sys.setrecursionlimit(10 ** 6)


def rolling_hash(s, w, MOD):
    ret = []
    tmp = 0
    p = pow(26, w, MOD)
    ords = [ord(c) - 97 for c in s]
    for i, o in enumerate(ords):
        tmp = tmp * 26 + o
        if i >= w:
            tmp = (tmp - ords[i - w] * p)
        tmp %= MOD
        ret.append(tmp)
    return ret


def solve(s, t):
    MOD = 10 ** 9 + 7
    ls, lt = len(s), len(t)
    k = (lt - 1) // ls + 1
    s *= k * 2
    ls *= k
    rs, rt = rolling_hash(s, lt, MOD), rolling_hash(t, lt, MOD)
    rs = rs[ls:]
    ht = rt[-1]

    checked = [-1] * ls

    def series(i, st):
        if i == st:
            return float('-inf')
        if checked[i] == -1:
            checked[i] = series((i + lt) % ls, st) + 1 if rs[i] == ht else 0
        return checked[i]

    for i, hs in enumerate(rs):
        if hs != ht:
            continue
        ret = series((i + lt) % ls, i)
        if ret == float('-inf'):
            return -1
        checked[i] = ret + 1

    return max(0, max(checked))


s = input()
t = input()
print(solve(s, t))