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 <algorithm>
#include <cmath>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <string.h>
#include <stack>
#define Endl endl
#define mp make_pair
#define ll long long 
#define pii pair<int,int>
#define pll pair<ll,ll>
#define over(A) {cout<<A<<endl;exit(0);}
#define all(A) A.begin(),A.end()
#define ceil(a,b) ((a-1)/b+1)
#define srand() mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rand(l,r) uniform_int_distribution<int>(l,r)(rng)
typedef unsigned long long ull;
const int inf=1039074182;
using namespace std;
char a[500005*6];
char b[500005*6];
int fail[500005*6];
int la,lb;

void build(char c[])
{
	int j=fail[0]=-1;
	for(int i=1;i<=lb;i++)
	{
		while(j>=0 && c[j]!=c[i-1]) j=fail[j];
		fail[i]=++j;
	}
}

int search(int l,int r)
{
	int res=0;
	int cur=0;
	for(int i=l;i<r;i++)
	{
		while(cur>=0 && a[i]!=b[cur]) cur=fail[cur];
		cur++;
		res=max(res,cur/lb);
	}
	return res;
}

int main()
{
//	freopen("input.txt","r",stdin);
	scanf("%s",a);
	scanf("%s",b);
	la=strlen(a);
	lb=strlen(b);
	int time=0;
	while(time*la<lb)
	{
		time++;
	}
	for(int i=1;i<time;i++)
	{
		for(int pos=0;pos<la;pos++)
		{
			a[i*la+pos]=a[pos];
		}
	}
	la=strlen(a);
	for(int i=1;i<3;i++)
	{
		for(int pos=0;pos<la;pos++)
		{
			a[i*la+pos]=a[pos];
		}
	}
	int timeb=0;
	while(timeb*lb<la*3)
	{
		timeb++;
	}
	for(int i=0;i<timeb;i++)
	{
		for(int pos=0;pos<lb;pos++)
		{
			b[i*lb+pos]=b[pos];
		}
	}
	build(b);
	int c1=search(0,la),c2=search(0,la*2),c3=search(0,la*3);
	if(c2==c1) over(c1);
	if(c2>c1 && c3>c2) over(-1);
	cout<<c2<<endl;
	return 0;
}