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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#define int long long
#include "string"

vector<int>preffix(string s) {
    int n=s.size();
    vector<int>pi(n,0);
    for(int i=1;i<n;i++) {
        int j=pi[i-1];
        while(j>0 and s[i]!=s[j])
            j=pi[j-1];
        if(s[i]==s[j])
            pi[i]=j+1;
    }
    return pi;
}

int kmp(string t,string p) {
    vector<int>ans;
    int n=t.size();
    int m=p.size();
    vector<int>pi=preffix(p);
    int j=0;
    for(int i=0;i<n;i++) {
        while(j>0 and t[i]!=p[j])
            j=pi[j-1];
        if(t[i]==p[j]) {
            j++;
            if(j==m){
                ans.push_back(i-j+1);
                j = pi[j-1];
            }
        }
    }
    return ans.size();
}

int solve(string s,string t){
	int l=1,r = s.size()/t.size() + 1,ans = -1;

	while(l<=r){
		int m = (l + r)/2;
		string tt = "";
		for(int i=0;i<m;i++)
			tt += t;
		if(kmp(s,tt) > 0){
			ans = m;
			l = m + 1;
		}
		else
			r = m - 1;
	}
	return ans;
}

int32_t main(){
	boost;
	string S,T;
	cin>>S>>T;
	string s = S;
	while(s.size() <= S.size()  + T.size())
		s += S;

	int ans1 = solve(s,T);
	if(ans1 == -1){
		cout<<0<<endl;
		return 0;
	}

	s += s;
	
	int ans2 = solve(s,T);

	if(ans1 != ans2)
		cout<<-1<<endl;
	else
		cout<<ans2<<endl;
}