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: Runtime Error)

S = input()
W = input()
S *= 6


N = len(W)
def primeFactor(N):
    i, n, ret, d, sq = 2, N, {}, 2, 99
    while i <= sq:
        k = 0
        while n % i == 0: n, k, ret[i] = n//i, k+1, k+1
        if k > 0 or i == 97: sq = int(n**(1/2)+0.5)
        if i < 4: i = i * 2 - 1
        else: i, d = i+d, d^6
    if n > 1: ret[n] = 1
    return ret

def divisors(N):
    pf = primeFactor(N)
    ret = [1]
    for p in pf:
        ret_prev = ret
        ret = []
        for i in range(pf[p]+1):
            for r in ret_prev:
                ret.append(r * (p ** i))
    return sorted(ret)

D = divisors(N)
p = 1
for d in D:
    if W[:-d] == W[d:]:
        W = W[:d]
        p = N//d
        N = d
        break

W = W[:d]


T = [-1] * (len(W)+1)

ii = 2
jj = 0

T[0] = -1
T[1] = 0

while ii < len(W):
    if W[ii - 1] == W[jj]:
        T[ii] = jj + 1
        ii += 1
        jj += 1
    elif jj > 0:
        jj = T[jj]
    else:
        T[ii] = 0
        ii += 1

def KMP(i0):
    ret = -1
    
    m = 0
    i = 0

    while m + i < len(S) - i0:
        if W[i] == S[m + i + i0]:
            i += 1
            if i == len(W):
                return m
        else:
            m = m + i - T[i]
            if i > 0:
                i = T[i]
    return len(S) - i0

i = 0
c = 0
cmax = 0
while i < len(S):
    # print("i =", i)
    k = KMP(i)
    if k == len(S):
        break
    elif k == 0:
        c += 1
    else:
        c = 1
    cmax = max(cmax, c)
    i += k + len(W)
if cmax * len(W) * 3 > len(S) * 2:
    print(-1)
else:
    print(cmax//0)