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)

class RollingHash():
    def __init__(self, s):
        self.length = len(s)
        self.base1 = 1009
        self.base2 = 1013
        self.mod1 = 10**9 + 7
        self.mod2 = 999999937 
        self.hash1 = [0] * (self.length + 1)
        self.hash2 = [0] * (self.length + 1)
        self.pow1 = [1] * (self.length + 1)
        self.pow2 = [1] * (self.length + 1)

        for i in range(self.length):
            self.hash1[i+1] = (self.hash1[i] + ord(s[i])) * self.base1 % self.mod1
            self.hash2[i+1] = (self.hash2[i] + ord(s[i])) * self.base2 % self.mod2
            self.pow1[i+1] = self.pow1[i] * self.base1 % self.mod1
            self.pow2[i+1] = self.pow2[i] * self.base2 % self.mod2
        
    def get(self, l, r):
        t1 = self.hash1[r] - self.hash1[l] * self.pow1[r-l] % self.mod1
        t1 = (t1 + self.mod1) % self.mod1
        t2 = self.hash2[r] - self.hash2[l] * self.pow2[r-l] % self.mod2
        t2 = (t2 + self.mod2) % self.mod2
        return (t1, t2)
    
t_tmp = input().rstrip()
s = input().rstrip()

t = ""
while len(t) < len(s):
    t += t_tmp
t = t + t

RHt = RollingHash(t)
RHs = RollingHash(s)
w = len(s)
s_hash = RHs.get(0, w)

infinity = True
lent = len(t)
lens = len(s)
for i in range(max(len(t)+1, len(s)+1)):
    if t[i%lent] != s[i%lens]:
        infinity = False
        break
if infinity:
    print(-1)
    exit()

match = [0] * (len(t) - len(s) + 1)
# print(match, t, s)
for i in range(len(t) - len(s) + 1):
    if RHt.get(i, i+w) == s_hash:
        match[i] = 1
ans = max(match[:w])
for i in range(len(match) - w):
    if match[i+w] != 0:
        match[i+w] += match[i]
    ans = max(ans, match[i+w])

print(ans)