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 rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
const int INF = (1<<30)-1;
const long long LINF = (1LL<<62)-1;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

template<class T> vector<int> z_algorithm(const T& s) {
  vector<int> prefix(s.size());
  prefix[0] = s.size();
  int i = 1, len = 0;
  while (i < s.size()) {
    while (i + len < s.size() && s[i + len] == s[len]) ++len;
    prefix[i] = len;
    if (!len) { ++i; continue; }
    int j = 1;
    while (i + j < s.size() && prefix[j] < len - j) {
      prefix[i + j] = prefix[j];
      ++j;
    }
    i += j;
    len -= j;
  }
  return prefix;
}

class UnionFind {
  vector<int> par, sz;
public:
  UnionFind(int n) {
    par.resize(n);
    for (int i = 0; i < n; i++)
      par[i] = i;
    sz.resize(n, 1);
  }
  int root(int x) {
    if (par[x] == x)
      return x;
    return par[x] = root(par[x]);
  }
  bool same(int x, int y) {
    return root(x) == root(y);
  }
  int size(int x) {
    return sz[root(x)];
  }
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    if (sz[x] >= sz[y]) {
      sz[x] += sz[y];
      par[y] = x;
    } else {
      sz[y] += sz[x];
      par[x] = y;
    }
    return true;
  }
};

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s, t;
  cin >> s >> t;
  while (s.size() < t.size())
    s += s;
  auto len = z_algorithm(t+s+s);
  UnionFind uf(s.size());
  bool exist = false;
  rep(i, s.size()) {
    if (len[t.size()+i] >= t.size()) {
      exist = true;
      if (len[t.size()+(i+t.size())%s.size()] >= t.size() && !uf.unite(i, (i+t.size())%s.size())) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  if (!exist) {
    cout << 0 << endl;
    return 0;
  }
  int ans = 0;
  rep(i, s.size())
    chmax(ans, uf.size(i));
  cout << ans << endl;
  return 0;
}