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: Wrong Answer)

#include <bits/stdc++.h>
#define mp make_pair
#define F first
#define S second
#define ll long long
#define all(x) x.begin(),x.end()
using namespace std;
const int LIM = 1e7+3;
int lps[LIM+3], last = 0;
string s,ss, cop;
int sz1, sz2;
std::vector<int> Indeces;
bool isSubStrin(string s, string ss)
{
  string tem = ss+"#"+s;
  int i = 1, len = 0;
  bool Ok = false;
  while(i<tem.size())
  {
    if(tem[i] == tem[len])len++, lps[i++] = len;
    else {
      if(len)
        len = tem[len-1];
      else i++;
    }
    if(len == ss.size()){Ok = true; Indeces.push_back(i);}
  }
  return Ok;
}
 
void getString(string &Cur, int last, int to)
{
    while(last<to)
        Cur+=cop, last++;
    while(to<last)
        Cur.erase(Cur.size()-sz2, sz2), last--;
}

bool vis[LIM+3];

int main(){
    cin >> s >> ss;
    bool Is = true;
    sz1 = s.size(); sz2 = ss.size();
    cop = s;
    while(sz1 < sz2)
      s += cop, sz1 = s.size();
    s += s; sz1 = s.size();
    cop = ss;
    int last = 1;
    if(!isSubStrin(s,ss)){cout << 0 << endl;return 0;}
    int Ans = 0, cnt = 0;
    for(int i =0;i<Indeces.size();i++)
    {
        if(vis[i])continue;
        cnt = 1;
        int st = Indeces[i];
        while(true)
        {
            int ind = lower_bound(all(Indeces), st+sz2)-Indeces.begin();
            if(ind == Indeces.size() or Indeces[ind] != st+sz2)break;
            vis[ind] = true; st = Indeces[ind]; cnt++;
        }
        Ans = max(Ans, cnt);
    }
    s += s;
    getString(ss, last, sz1/sz2 + 1);
    if(isSubStrin(s,ss)){puts("-1");return 0;}
    cout << Ans << endl;
}