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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;

class AtCoder
{
    const int MOD = 1000000007;
    public void Slove()
    {
        var s = Console.ReadLine();
        var t = Console.ReadLine();
        var ans = 0;
        var count = 0;
        bool many = true;
        for(int i = 0; i < Math.Max(s.Length, t.Length) * 3; i++)
        {
            //Console.WriteLine(s[i % s.Length]);
            if (t[0] == s[i % s.Length])
            {
                if (Enumerable.Range(0, t.Length).All(x => t[x] == s[(i + x) % s.Length]))
                {
                    count++;
                    i += t.Length - 1;
                    if(ans < count)
                    {
                        ans = count;
                    }
                }
                else
                {
                    count = 0;
                    many = false;
                }
            }
            else
            {
                count = 0;
                many = false;
            }
        }
        Console.WriteLine(many ? -1 : ans);
    }
    static void Main()
    {
        var ac = new AtCoder();
        ac.Slove();
    }
}