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 <bits/stdc++.h>
using namespace std;
using lint = long long int;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
constexpr int MOD = 1000000007;
int main()
{
    string s, t;
    cin >> s >> t;
    string sloop = "";
    while(sloop.size() < s.size() + t.size()) {
        sloop += s;
    }
    vector<int> table(t.size(), 0);
    int i = 2;
    int j = 0;
    table[0] = -1;
    if(t.size() >= 2) table[1] = 0;
    while(i < t.size()) {
        if(t[i-1] == t[j]) {
            table[i] = j + 1;
            i++;
            j++;
        } else if(j > 0) {
            j = table[j];
        } else {
            table[i] = 0;
            i++;
        }
    }
    int m = 0;
    i = 0;
    vector<bool> pat(s.size(), false);
    while(m + i < sloop.size()) {
        if(t[i] == sloop[m + i]) {
            i++;
            if(i == t.size()) {
                pat[m] = true;
                i--;
                m = m + i - table[i];
                if(i > 0) {
                    i = table[i];
                }
            }
        } else {
            m = m + i - table[i];
            if(i > 0) {
                i = table[i];
            }
        }
    }
    int ans = 0;
    vector<bool> used(s.size(), false);
    int smul = 0;
    while(smul < t.size()) {
        smul += s.size();
    }
    REP(i, s.size()) {
        if(pat[i] == true && used[i] == false) {
            int j = i;
            while(true) {
                if(pat[(j - t.size() + smul) % s.size()] == true) {
                    j = (j - t.size() + smul) % s.size();
                    if(i == j) {
                        cout << -1 << endl;
                        return 0;
                    }
                } else {
                    break;
                }
            }
            int k = 0;
            while(true) {
                used[j] = true;
                k++;
                if(pat[(j + t.size()) % s.size()] == true) {
                    j = (j + t.size()) % s.size();
                } else {
                    break;
                }
            }
            if(k > ans) ans = k;
        }
    }

    cout << ans << endl;
}