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: Time Limit Exceeded)

#define DEBUG 0
/**
 * File    : F.cpp
 * Author  : Kazune Takahashi
 * Created : 7/28/2019, 1:18:49 AM
 * Powered by Visual Studio Code
 */
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <string>
#include <complex>
#include <tuple>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <functional>
#include <random>
#include <chrono>
#include <cctype>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define maxs(x, y) (x = max(x, y))
#define mins(x, y) (x = min(x, y))
using ll = long long;
class mint
{
public:
  static ll MOD;
  ll x;
  mint() : x(0) {}
  mint(ll x) : x(x % MOD) {}
  mint operator-() const { return x ? MOD - x : 0; }
  mint &operator+=(const mint &a)
  {
    if ((x += a.x) >= MOD)
    {
      x -= MOD;
    }
    return *this;
  }
  mint &operator-=(const mint &a) { return *this += -a; }
  mint &operator*=(const mint &a)
  {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint &operator/=(const mint &a)
  {
    mint b{a};
    return *this *= b.power(MOD - 2);
  }
  mint operator+(const mint &a) const { return mint(*this) += a; }
  mint operator-(const mint &a) const { return mint(*this) -= a; }
  mint operator*(const mint &a) const { return mint(*this) *= a; }
  mint operator/(const mint &a) const { return mint(*this) /= a; }
  bool operator<(const mint &a) const { return x < a.x; }
  bool operator==(const mint &a) const { return x == a.x; }
  const mint power(ll N)
  {
    if (N == 0)
    {
      return 1;
    }
    else if (N % 2 == 1)
    {
      return *this * power(N - 1);
    }
    else
    {
      mint half = power(N / 2);
      return half * half;
    }
  }
};
ll mint::MOD = 1e9 + 7;
istream &operator>>(istream &stream, mint &a) { return stream >> a.x; }
ostream &operator<<(ostream &stream, const mint &a) { return stream << a.x; }
class combination
{
public:
  vector<mint> inv, fact, factinv;
  static int MAX_SIZE;
  combination() : inv(MAX_SIZE), fact(MAX_SIZE), factinv(MAX_SIZE)
  {
    inv[1] = 1;
    for (auto i = 2; i < MAX_SIZE; i++)
    {
      inv[i] = (-inv[mint::MOD % i]) * (mint::MOD / i);
    }
    fact[0] = factinv[0] = 1;
    for (auto i = 1; i < MAX_SIZE; i++)
    {
      fact[i] = mint(i) * fact[i - 1];
      factinv[i] = inv[i] * factinv[i - 1];
    }
  }
  mint operator()(int n, int k)
  {
    if (n >= 0 && k >= 0 && n - k >= 0)
    {
      return fact[n] * factinv[k] * factinv[n - k];
    }
    return 0;
  }
};
int combination::MAX_SIZE = 3000010;
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
// constexpr double epsilon = 1e-10;
// constexpr ll infty = 1000000000000000LL;
// constexpr int dx[4] = {1, 0, -1, 0};
// constexpr int dy[4] = {0, 1, 0, -1};
void Yes()
{
  cout << "Yes" << endl;
  exit(0);
}
void No()
{
  cout << "-1" << endl;
  exit(0);
}

class UnionFind
{
  vector<long long> par;

public:
  UnionFind() {}
  UnionFind(int n) : par(n, -1) {}

  bool is_same(int x, int y)
  {
    return root(x) == root(y);
  }

  bool merge(int x, int y)
  {
    x = root(x);
    y = root(y);
    if (x == y)
    {
      return false;
    }
    if (par[x] > par[y])
    {
      swap(x, y);
    }
    par[x] += par[y];
    par[y] = x;
    return true;
  }

  long long size(int x)
  {
    return -par[root(x)];
  }

private:
  int root(int x)
  {
    if (par[x] < 0)
    {
      return x;
    }
    return par[x] = root(par[x]);
  }
};

int main()
{
  string S, T;
  cin >> S >> T;
  int N = S.size();
  stringstream U{};
  U << S << S << S;
  while (U.str().size() < T.size() * 2)
  {
    U << S;
  }
  S = U.str();
  vector<bool> ok(N, false);
  for (int i = 0; i < N; i++)
  {
    if (S.substr(i, T.size()) == T)
    {
      ok[i] = true;
    }
  }
  UnionFind uf{N};
  for (auto i = 0; i < N; i++)
  {
    int j = (i + T.size()) % N;
    if (ok[i] && ok[j])
    {
      if (!uf.merge(i, j))
      {
        No();
      }
    }
  }
  ll ans = 0;
  for (auto i = 0; i < N; i++)
  {
    if (ok[i])
    {
      maxs(ans, uf.size(i));
    }
  }
  cout << ans << endl;
}