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()
T = input()

if len(set(S))==1 and len(set(T))==1:
    print(-1)
    exit()
if len(T)%len(S) == 0:
    if S in T:
        i = T.index(S)
        TT = T + T
        for j in range(len(T)//len(S)):
            if TT[i+(j+1)*len(S):i+(j+2)*len(S)] != S:
                break
        else:
            print(-1)
            exit()
if len(S)%len(T) == 0:
    if T in S:
        i = S.index(T)
        SS = S + S
        for j in range(len(S)//len(T)):
            if SS[i+(j+1)*len(T):i+(j+2)*len(T)] != T:
                break
        else:
            print(-1)
            exit()

A = S * (1234567 // len(S))
assert len(A) > len(T)*2

if T not in A:
    print(0)
    exit()

i = seq = ans = 0
for a in A:
    if a == T[i]:
        i += 1
        if i==len(T):
            seq += 1
            ans = max(ans, seq)
            i = 0
    else:
        i = seq = 0
print(ans)