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: Wrong Answer)

s = input()
t = input()

def z_algorithm(s):
    n = len(s)
    res = [0] * n
    i = 1
    j = 0
    while i < n:
        # i番目以降の一致文字数
        while i + j < n and s[j] == s[i + j]:
            j += 1
        res[i] = j

        # 一文字も一致しない場合，次の文字へ
        if j == 0:
            i += 1
            continue

        # 一致したところまでを埋める
        k = 1
        while i + k < n and k + res[k] < j:
            res[i + k] = res[k]
            k += 1
        
        i += k
        j -= k
        
    return res


n = len(t)
if n > len(s):
    ss = s * (n // len(s) + 1)
else:
    ss = s

z = z_algorithm(t + ss * 2)
z = [min(z[i], n) for i in range(n, len(z))]

ans = 0
for i in range(n):
    c = 0
    for v in z[i::n]:
        if v == n:
            c += 1
        else:
            c = 0
        ans = max(ans, c)
        
if ans <= len(ss) // n:
    print(ans)
else:
    print(-1)