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)

import sys
sys.setrecursionlimit(10**9)
s = input()
t = input()
ls = len(s)
lt = len(t)
S = s*(lt//ls+2)
N = len(S)
mod = 2**61-1
base = 4#ランダムに撮るのが好ましい
power = [1]*N#power[i] = base^iの配列
for i in range(1, N):
    power[i] = power[i-1]*base%mod
rolling_hash = [0]*(N+1)
k = lt
for i, s in enumerate(S[:k]):
    rolling_hash[i+1] = rolling_hash[i]*base+ord(s)
for i in range(k, N):
    rolling_hash[i+1] = ((rolling_hash[i]-ord(S[i-k])*power[k-1])*base+ord(S[i]))%mod
thash = sum(ord(t[i])*power[lt-i-1] for i in range(lt))
is_same = [False]*ls
for i, h in enumerate(rolling_hash[k:ls+k]):
    if h == thash:
        is_same[i] = True
length = [-1]*N
is_visited = [False]*N
def rec(i):
    if length[i]>=0:
        return length[i]
    if is_visited[i]:
        return -1
    is_visited[i] = True
    if not is_same[i]:
        length[i] = 0
        return 0
    ni = (i+lt)%ls
    length[i] = rec(ni)+1
    if length[i] == 0:
        return -1
    return length[i]
for i in range(ls):
    if is_visited[i]:
        continue
    res = rec(i)
    if res == -1:
        ans = -1
        break
    ans = max(ans, res)
print(ans)