Problem:
Score : 300 points

Problem Statement

You are given a string S of length N and another string T of length M.
These strings consist of lowercase English letters.

A string X is called a good string when the following conditions are all met:

Let L be the length of X. L is divisible by both N and M.

Concatenating the 1-st, (\frac{L}{N}+1)-th, (2 \times \frac{L}{N}+1)-th, ..., ((N-1)\times\frac{L}{N}+1)-th characters of X, without changing the order, results in S.

Concatenating the 1-st, (\frac{L}{M}+1)-th, (2 \times \frac{L}{M}+1)-th, ..., ((M-1)\times\frac{L}{M}+1)-th characters of X, without changing the order, results in T.

Determine if there exists a good string. If it exists, find the length of the shortest such string.

Constraints

1 \leq N,M \leq 10^5

S and T consist of lowercase English letters.

|S|=N

|T|=M

Input

Input is given from Standard Input in the following format:

N M
S
T

Output

If a good string does not exist, print -1; if it exists, print the length of the shortest such string.

Sample Input 1

3 2
acp
ae

Sample Output 1

6

For example, the string accept is a good string.
There is no good string shorter than this, so the answer is 6.

Sample Input 2

6 3
abcdef
abc

Sample Output 2

-1

Sample Input 3

15 9
dnsusrayukuaiia
dujrunuma

Sample Output 3

45

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        String S = sc.next();
        String T = sc.next();

        if (N % M == 0 || M % N == 0) {
            if (N > 26 || M > 26) {
                System.out.println(-1);
                return;
            }
            if (S.contains(T) || T.contains(S)) {
                System.out.println(-1);
                return;
            }
            int[] X = new int[26];
            for (int i = 0; i < N; i++) {
                X[S.charAt(i)] = S.charAt(i) - 'a';
            }
            for (int i = 0; i < M; i++) {
                if (X[T.charAt(i)] != 0) {
                    System.out.println(-1);
                    return;
                }
            }
        } else {
            System.out.println(lcm(N, M));
        }
        sc.close();

    }
    static long lcm (int a, int b) {
        int temp;
        long c = a;
        c *= b;
        while((temp = a % b) != 0) {
            a = b;
            b = temp;
        }
        return c / b;
    }

}