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 (GCC 5.4.1) (judge verdict: Runtime Error)

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<stdio.h>

int v[1500001];
char s[1500001],t[500002];
void Zalgorithm(int N){
  int i=1,j=0,k=0;
  while(i<N){
    while(i+j<N&&s[i+j]==s[j])++j;
    v[i]=j;
    if(!j){
      ++i;
      continue;
    }
    k=1;
    while(v[k]+k<j){
      v[i+k]=v[k];
      ++k;
    }
    i+=k;
    j-=k;
  }
}
int i,j,K,que[500001],bf,ans,size;
int main(){
  int N,ng=500001,mid;
  gets(t);
  gets(s);
  while(ng-K>1){
    mid=(ng+K)>>1;
    if(s[mid])K=mid;
    else ng=mid;
  }
  while(t[i]!='\n'){
    s[K+i]=t[i];++i;
  }
  N=i;
  i+=K;
  int M=N+(K<<1);
  while(i<M){
    s[i++]=t[j++];
    j*=j<N;
  }
  Zalgorithm(M);
  for(i=0;i<N;++i){
    if(v[i+K]<K)que[size++]=(i+K)%N;
  }
  for(i=0;i<N;++i){
    if(i==size){
      puts("-1");
      return 0;
    }
    if(v[que[i]+=K]>=K){
      que[size++]=que[i]%N;
      if(bf<=i){
        bf=size-1;
        ++ans;
      }
    }
  }
  printf("%d\n",ans);
}