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: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=4000002;
char s[N],t[N];
int a,b,ne[N],ans,g[N],i,j;
int main(){
	scanf("%s%s",s+1,t+1),a=strlen(s+1),b=strlen(t+1);
	for (;a<b*2;a*=2)
		for (i=1;i<=a;i++) s[i+a]=s[i];
	for (i=1;i<=a;i++) s[i+a]=s[i];
	a*=2;
//	for (i=2,j=1;i<=b;i++){
//		while (j>1 && t[i]!=t[j]) j=ne[j-1];
//		if (t[i]==t[j]) j++;
//		ne[i]=j;
//	}
	ne[0]=-1;
	for (i=1;i<=b;i++){
		j=i-1;
		while (j && t[ne[j]+1]!=t[i]) j=ne[j];
		ne[i]=ne[j]+1;
	}
//	for (i=1,j=1;i<=a;i++){
//		while (j>1 && s[i]!=t[j]) j=ne[j-1];
//		if (s[i]==t[j]) j++;
//		if (j==b+1) ans=max(ans,g[i]=g[i-b]+1);
//	}
	for (i=1,j=0;i<=a;i++){
		while (~j && s[i]!=t[j+1]) j=ne[j];
		if (++j==b) ans=max(ans,g[i]=g[i-b]+1);
	}
	printf("%d",ans>=a/b-1?-1:ans);
}