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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define For(i,a,b) for(int i=a;i<=b;i++)
#define F first
#define S second
#define all(x) x.begin(),x.end()
#define sz(x) ((int)x.size())
#define MOD (ll)(INF)
#define INF 1e17
#define int ll
using namespace std;
using ll=long long;
using pii=pair<int,int>;

int gcd(int a,int b) { return b==0?a:gcd(b,a%b); }
int lcm(int a,int b) { return a/gcd(a,b)*b; }

int fpow(int b,int p)
{
	if(p==0) return 1;
	int h=fpow(b,p/2);
	if(p%2) return h*b%MOD*h%MOD;
	return h*h%MOD;
}

int32_t main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	string a,b;
	cin>>a>>b;
	For(i,0,max(sz(a),sz(b))-1){
		if(a[i%sz(a)]!=b[i%sz(b)]) goto END;
	}
	cout<<"-1\n";
	return 0;
END:
	int cnt=0;
	int i;
	for(i=0;i<sz(a);i++)
		if(a[i]!=b[i%sz(b)]) break;
	cnt+=i/sz(b);
	//cout<<i<<" "<<cnt<<"\n";
	for(i=0;i<sz(a);i++)
		if(a[sz(a)-i-1]!=b[sz(b)-(i%sz(b))-1]) break;
	cnt+=i/sz(b);
	//cout<<i<<" "<<cnt<<"\n";
	cout<<cnt<<"\n";
	return 0;
}