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 <bits/stdc++.h>


#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define MOD 1000000007
#define pii pair<int,int>
#define ll long long
#define el '\n'

using namespace std;

bool cek[3000100];

vector <int> prefix(string s){
    int n=s.length();
    vector <int> ans(n);

    int l=0;
    ans[0]=0;

    for (int r=1;r<n;r++){
        while (l!=0 && s[l]!=s[r]) l=ans[l-1];

        if (s[l]==s[r]){
            ans[r]=l+1;
            l++;
        }
        else ans[r]=0;
    }

    return ans;
}

int main()
{

    string s,t;
    cin >> s >> t;

    string st=s;

    for (int i=0;i<st.length();i++){
        s.pb(st[i]);
    }
    for (int i=0;i<st.length();i++){
        s.pb(st[i]);
    }
    for (int i=0;i<st.length();i++){
        s.pb(st[i]);
    }
    while (s.length()<t.length()){
        for (int i=0;i<st.length();i++){
            s.pb(st[i]);
        }
    }

    vector <int> pi_t,ans;
    pi_t=prefix(t);
    int n=s.length(),m=t.length();
    int ps=0,pt=0,cnt=0;
 //   cerr << n << " " << m << el;

    while (ps<n){
        while (pt!=0 && s[ps]!=t[pt]) pt=pi_t[pt-1];

        if (s[ps]==t[pt]) pt++;
        ps++;

        if (pt>=m){
        //    ans.push_back(ps-m+1);
            cek[ps-m]=1;
            cnt++;
        }
    }


    int res=0;
    bool infin=0;
    bool semua=1;
    for (int i=0;i<m;i++){
        int tm=0;
        bool bef=0;
        bool inf=1;
        for (int j=i;j<n-m+1;j+=m){
            inf=(inf&&cek[j]);
            semua=(semua&&cek[j]);
            if (bef){
                if (cek[j]){
                    tm++;
                } else{
                    tm=0;
                    bef=0;
                }
            } else{
                if (cek[j]){
                    tm=1;
                    bef=1;
                }
            }
            res=max(res,tm);
        }
        if (inf && (n%m==0)) infin=1;
    }

    if (semua||infin){
        cout << -1 << el;
        return 0;
    }
    cout << res << el;
    
}