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: Runtime Error)

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100000;

char A[maxn+5],B[maxn+5];
int n,m,pre,suf,pos[maxn+5];
int ans,P,S;

int main()
{
    scanf("%s%s",A+1,B+1);n=strlen(A+1);m=strlen(B+1);
    int j=n;suf=m;
    while (suf>=1)
    {
        while (j>=1&&A[j]!=B[suf]) j--;if (j<1) break;
        pos[suf]=j;suf--;j--;
    }
    suf++;if (m-suf+1>ans) ans=m-suf+1,P=0,S=suf;
    pre=1;j=1;
    while (pre<=m)
    {
        while (j<=n&&A[j]!=B[pre]) j++;if (j>n) break;
        while (suf<=m&&suf<=pre) suf++;
        while (suf<=m&&pos[suf]<=j) suf++;
        if (pre+m-suf+1>ans) ans=pre+m-suf+1,P=pre,S=suf;
        pre++;j++;
    }
    if (!ans) return puts("-"),0;
    for (int i=1;i<=P;i++) putchar(B[i]);
    for (int i=S;i<=m;i++) putchar(B[i]);
    return putchar('\n'),0;
}