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 <iostream>
#include <cstring>
#include <map>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;


int a[2100000];
int check[510000],vis[510000];

int getans(int i,int ss,int tt)
{
	int cnt=0,i0=i;
	while(check[i])
	{
		if (vis[i]) return -1;
		cnt++;
		vis[i]=1;
		i=i-tt;
		if (i<0) i+=ss;	
	}
	i=(i0+tt)%ss;
	while(check[i])
	{
		if (vis[i]) return -1;
		cnt++;
		vis[i]=1;
		i=(i+tt)%ss;	
	}
	return cnt;
}
int main()
{
    string s,t,s0; cin>>s>>t; s0=s;
    int ss=s.length(),tt=t.length();
	while(s0.length()<ss+tt) s0=s0+s;
	s0=t+s0;
	a[0]=-1; int k=0;
	for (int i=1; i<s0.length(); i++)
	{
		while(s0[i]!=t[k]&&k!=-1) 
		{
			if (k>0) k=a[k-1];
			else
			{
				k=-1; break;
			}
		}
		k++; if (k>tt) k=a[k-1];		
		a[i]=k;
		if (k==tt) check[(i-tt)%ss]=1; 
	}
	
	int ans=0;
	for (int i=0; i<ss; i++)
	    if (check[i]&&!vis[i])
		{
			int xx=getans(i,ss,tt);
			if (xx==-1)
			{
				cout << -1;
				return 0;
			}
			ans=max(ans,xx);
		} 
	cout << ans;
}