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: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
int max(const void *a, const void *b){return *(int*)b - *(int*)a;}

int kmp[500001]={0};
void getKMP(char* str, int len)
{
    int i = 1;
    int j = 0;

    while(str[i]!='\0')
    {
        if(str[i] == str[j]){
            j++;
            kmp[i] = j;
            i++;
        }else if(j==0)
        {
            i++;
        }else{
            j = kmp[j-1];
        }
    }
}

//Union Find
int uf[500001];
int ufSize[500001];
void UFinit(int n)
{
    for(int i=0;i<n;i++)
    {
        uf[i] = i;
        ufSize[i] = 1;
    }
}

int UFroot(int i)
{
    if(uf[i]==i) return i;
    uf[i] = UFroot(uf[i]);
    return uf[i];
}

// 元々根が同じならfalse 違うなら統合してtrue
bool UFunite(int i,int j)
{
    int ri = UFroot(i);
    int rj = UFroot(j);

    if(ri == rj) return false;
    ufSize[ri] += ufSize[rj];
    uf[rj] = ri;
    return true;
}

char s[500001];
char t[500001];
bool match[500001];
size_t sl,tl;

int main(void)
{
    scanf("%s",s);
    scanf("%s",t);
    sl = strlen(s);
    tl = strlen(t);
    getKMP(t,tl);
    // for(int i=0;i<tl;i++) printf("%d\n",kmp[i]);
    {
        int i = 0;
        int j = 0;
        while(i<sl+tl)
        {
            if(s[i%sl]==t[j])
            {
                i++;
                j++;
                if(j==tl)
                {
                    match[i-tl] = true;
                    j = kmp[tl-1];
                }
            }else
            {
                if(j==0){
                    i++;
                    j = 0;
                }else
                {
                    j = kmp[j-1];
                }
                
                
            }
        //    printf("%d %d\n",i,j);
        }
    }

//    for(int i=0;i<sl;i++) printf("%d\n",match[i]);
    UFinit(sl);
    for(int i=0;i<sl;i++)
    {
        if(match[i])
        {
            if(!UFunite(i,(i+tl)%sl))
            {
                printf("-1\n");
                return 0;
            }
        }
    }
    int max;
    max = 0;
    for(int i=0;i<sl;i++) max = ufSize[i] > max ? ufSize[i] : max;
    printf("%d\n", max-1);
    
    return 0;
}