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;

vector<int> Z(string s) {
    vector<int> z(s.size() , 0);
    int l = 0 , r = 0;
    for (int i = 1 ;i < s.size() ;i++) {
        if (i <= r)
            z[i] = min(z[i - l] , r - i + 1);

        while (i + z[i] < s.size() && s[i + z[i]] == s[z[i]])
            z[i]++;

        if (i + z[i] - 1 > r)
            l = i , r = i + z[i] - 1;
    }

    return z;
}

const int N = 5e5 + 5;
vector<int> adj[N];
int vis[N];

bool cycle(int node) {
    if (vis[node])
        return vis[node] == 1;

    vis[node] = 1;
    for (int c : adj[node])
        if (cycle(c))
            return 1;
    vis[node] = 2;

    return 0;
}

int memo[N];
int solve (int node) {
    int &ret = memo[node];
    if (~ret)
        return ret;

    ret = 0;
    for (int c : adj[node])
        ret = max(ret , solve(c) + 1);

    return ret;
}

int main() {
    ios::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);

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

    while (s.size() < t.size())
        s += s;

    int n = s.size();
    int m = t.size();

    vector<int> z = Z(t + "$" + s + s);
    int st = m + 1;
    int en = st + n - 1;

    for (int i = st ;i <= en ;i++)
        if (z[i] >= m)
            adj[i - st].push_back((i - st + m) % n);

    for (int i = 0 ;i < s.size() ;i++) {
        if (cycle(i)) {
            cout << -1;
            return 0;
        }
    }

    memset(memo , -1 , sizeof memo);
    int mx = 0;
    for (int i = 0 ;i < s.size() ;i++)
        mx = max(mx , solve(i));

    cout << mx;
}