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: Accepted)

#include <bits/stdc++.h>
#define all(vec) vec.begin(), vec.end()
#define pb push_back
#define eb emplace_back
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
constexpr ll INF = (1LL << 30) - 1LL;
constexpr ll MOD = 1e9 + 7;
constexpr int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
template <class T>
void chmin(T &a, T b) { a = min(a, b); }
template <class T>
void chmax(T &a, T b) { a = max(a, b); }
void ok() { cerr << "ok" << endl; }
using ull = uint64_t;
const int bases[64] = {257, 262, 266, 275, 276, 281, 285, 290, 296, 302, 306,
                       310, 311, 313, 323, 333, 344, 345, 350, 357, 367, 370,
                       373, 402, 423, 425, 431, 440, 442, 443, 454, 457, 458,
                       462, 471, 478, 481, 487, 489, 492, 499, 501, 502, 503,
                       506, 514, 524, 532, 535, 541, 550, 552, 557, 559, 562,
                       563, 567, 570, 571, 580, 592, 597, 604, 612};
const ull rmod = 0x1fffffffffffffff;
const ull base = bases[chrono::duration_cast<chrono::microseconds>(chrono::system_clock::now().time_since_epoch()).count() & 63];
const ull mask30 = (1UL << 30) - 1, mask31 = (1UL << 31) - 1;
struct RollingHash {
    vector<ull> hash, pow;
    inline ull mul(const ull &a, const ull &b) const {
        ull mid = (a & mask31) * (b >> 31) + (a >> 31) * (b & mask31);
        ull res = (a >> 31) * (b >> 31) * 2 + (mid >> 30) + ((mid & mask30) << 31);
        res += (a & mask31) * (b & mask31);
        res = (res >> 61) + (res & rmod);
        if (res >= rmod) res -= rmod;
        return res;
    }
    RollingHash(const string &s) {
        hash.resize(s.length() + 1);
        pow.resize(s.length() + 1);
        pow[0] = 1;
        for (int i = 0; i < s.length(); i++) {
            hash[i + 1] = mul(hash[i], base) + s[i];
            pow[i + 1] = mul(pow[i], base);
            if (hash[i + 1] >= rmod) hash[i + 1] -= rmod;
        }
    }
    ull get(int l, int r) { //[l,r]
        ull res = hash[r + 1] + rmod - mul(hash[l], pow[r + 1 - l]);
        if (res >= rmod) res -= rmod;
        return res;
    }
};
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s, t;
    cin >> s >> t;
    int n = s.length(), m = t.length();
    while (s.length() < n + m) {
        s += s;
    }
    RollingHash hs(s), ht(t);
    vector<int> f(n), vis(n, -1);
    for (int i = 0; i < n; i++) {
        if (hs.get(i, i + m - 1) == ht.get(0, m - 1)) {
            f[i] = 1;
        }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        if (!f[i] || vis[i] != -1) {
            continue;
        }
        int co = 0, id = i;
        vector<int> v;
        while (1) {
            v.push_back(id);
            co++;
            id += m;
            id %= n;
            if (f[id]) {
                if (id == i) {
                    cout << -1 << endl;
                    return 0;
                }
                if (vis[id] != -1) {
                    co += vis[id];
                    break;
                }
                vis[id] = 0;
            } else {
                break;
            }
        }
        vis[i] = co;
        chmax(ans, co);
    }
    cout << ans << endl;
}