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<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define ll long long
#define ull unsigned long long
#define maxn 1000010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static char buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10+'0';
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]);
}
inline void writesp(ll x){write(x); putchar(' ');}
inline void writeln(ll x){write(x); putchar('\n');}
char s[maxn],t[maxn];
int f[maxn];
ull a[maxn],b[maxn],p[maxn];
int n,m;
ull getHash_s(int l,int r){return a[r]-a[l-1]*p[r-l+1];}
ull getHash_t(int l,int r){return b[r]-b[l-1]*p[r-l+1];}
int main()
{
	scanf("%s",s+1); scanf("%s",t+1);
	n=strlen(s+1); m=strlen(t+1);
	p[0]=1; a[0]=b[0]=0;
	for(int i=1;i<=2*n||i<=m;i++)p[i]=p[i-1]*233;
	for(int i=1;i<=n;i++)a[i]=a[i-1]*233+s[i];
	for(int i=1;i<=m;i++)b[i]=b[i-1]*233+t[i];
	int lens=n;
	for(int i=1;i<=n;i++)
		if(n%i==0){
			int flag=1;
			for(int j=2*i;j<=n;j+=i)
				if(getHash_s(j-2*i+1,j-i)!=getHash_s(j-i+1,j)){
					flag=0; break;
				}
			if(flag){
				lens=i; break;
			}
		}
	int lent=m;
	for(int i=1;i<=m;i++)
		if(m%i==0){
			int flag=1;
			for(int j=2*i;j<=m;j+=i)
				if(getHash_t(j-2*i+1,j-i)!=getHash_t(j-i+1,j)){
					flag=0; break;
				}
			if(flag){
				lent=i; break;
			}
		}
	for(int i=lens+1;i<=2*lens;i++){
		s[i]=s[i-lens];
		a[i]=a[i-1]*233+s[i];
	}
	if(lens==lent){
		int flag=0;
		for(int i=1;i<=lens;i++)
			if(getHash_t(1,lent)==getHash_s(i,i+lens-1)){
				flag=1; break;
			}
		if(flag){
			puts("-1"); return 0;
		}
	}
	int ans=0;
	for(int i=m;i<=2*lens;i++)
		if(getHash_t(1,m)==getHash_s(i-m+1,i)){
			f[i]=f[i-m]+1;
			ans=std::max(ans,f[i]);
		}
	writeln(ans);
	return 0;
}