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;

class F
{
	static void Main()
	{
		var s = Console.ReadLine();
		var t = Console.ReadLine();

		var t_sub = new int[t.Length];
		t_sub[0] = -1;
		for (int i = 1; i < t.Length; i++)
		{
			var p = t_sub[i - 1];
			t_sub[i] = p > -1 && t[i] == t[p + 1] ? p + 1 : t[i] == t[0] ? 0 : -1;
		}

		var st = new bool[s.Length];
		for (int i = 0, j = 0; i < s.Length;)
		{
			for (; ; j++)
			{
				if (s[(i + j) % s.Length] != t[j]) break;
				if (j == t.Length - 1) { st[i] = true; break; }
			}

			if (t_sub[j] == -1)
			{
				i += j + 1;
				j = 0;
			}
			else
			{
				i += j - t_sub[j];
				j = t_sub[j] + 1;
			}
		}

		var M = 0;
		var u = new int[s.Length];
		for (int i = 0; i < s.Length; i++)
		{
			if (!st[i] || u[i] > 0) continue;

			var j = i;
			while (st[MInt(j - t.Length, s.Length)])
			{
				j = MInt(j - t.Length, s.Length);
				if (j == i) { Console.WriteLine(-1); return; }
			}

			var c = 0;
			while (st[j])
			{
				u[j] = ++c;
				j = (j + t.Length) % s.Length;
			}
			M = Math.Max(M, c);
		}
		Console.WriteLine(M);
	}

	static int MInt(int x, int mod) => (x %= mod) < 0 ? x + mod : x;
}