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: Wrong Answer)

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long n = sc.nextInt();
		long m = sc.nextInt();
		String s = sc.next();
		String t = sc.next();
		long l = lcm(n, m);
		Map<Long, String> map = new HashMap<>();
		for (int i = 0; i < n-1; i++) {
			map.put((long) (i*(l/n)+1), String.valueOf(s.charAt(i)));
		}
		for (int i = 0; i < m-1; i++) {
			if (map.containsKey((long) (i*(l/m)+1))) {
				if (!map.get((long) ((i*(l/m)+1))).equals(String.valueOf(t.charAt(i)))) {
					System.out.println(-1);
					return;
				}
			}
		}
		System.out.println(l);
	}
	
	private static long lcm(long n, long m) {
		long temp;
		long c = n;
		c *= m;
		while((temp = n%m) != 0) {
			n = m;
			m = temp;
		}
		return c / m;
	}
}