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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String S = in.nextLine();
        String T = in.nextLine();
        if(S.length() < T.length()) {
            int c = T.length() / S.length();
            if (T.length() % S.length() != 0) {
                c++;
            }
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < c; i++) {
                sb.append(S);
            }
            S = sb.toString();
        }
        String S2 = S.concat(S);
        if (!S2.contains(T)) {
            System.out.println(0);
        } else {
            S2 = S2.replace(T, "A");
            int count = 0;
            int max = 0;
            for (int i = 0; i < S2.length(); i++) {
                if (S2.charAt(i) == 'A') {
                    count++;
                } else {
                    if (count > max) {
                        max = count;
                    }
                    count = 0;
                }
                if (count == S.length() / T.length() + 1 || (count == S.length() / T.length() && S.length() % T.length() == 0)) {
                    max = -1;
                    break;
                }
            }
            System.out.println(max);
        }
    }
}