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: Accepted)



import java.util.Scanner;

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

		if (S.charAt(0) != T.charAt(0)) {
			System.out.println(-1);
			return;
		}

		int g = gcd(N, M);
		if (g == 1) {
			System.out.println((long) M * N / g);
		} else {
			int n = N / g;
			int m = M / g;
			//System.out.println(S.charAt(m));
			//System.out.println(T.charAt(n));
			int i = 1;
			while (n < N && m < M) {
				if (S.charAt(n) != T.charAt(m)) {
					System.out.println(-1);
					return;
				}
				i++;
				n = (N / g) * i;
				m = (M / g) * i;
			}
			System.out.println((long) M * N / g);

		}

	}

	private static int gcd(int n, int m) {
		int r;
		while ((r = n % m) != 0) {
			n = m;
			m = r;
		}
		return m;
	}
}