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;

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();
		scan.close();
		if(S.charAt(0) != T.charAt(0)) {
			System.out.println(-1);
			System.exit(0);
		}
		if(N == M) {
			if(S.equals(T)) {
				System.out.println(N);
			}else {
				System.out.println(-1);
			}
			System.exit(0);
		}
		int L = lcm(N, M);
		Map<Integer, Integer> map1 = new HashMap<Integer, Integer>();
		Map<Integer, Integer> map2 = new HashMap<Integer, Integer>();
		for(int i = 1; i < N; i++) {
			int t = i * L / N + 1;
			map1.put(t, i);
		}
		for(int i = 1; i < M; i++) {
			int t = i * L / M + 1;
			map2.put(t, i);
		}

		if(N % M == 0 || M % N == 0) {
			for(int k : map1.keySet()) {
				if(map2.containsKey(k)) {
					int t1 = map1.get(k);
					int t2 = map2.get(k);
					if(S.charAt(t1) != T.charAt(t2)) {
						System.out.println(-1);
						System.exit(0);
					}
				}
			}
			for(int k : map2.keySet()) {
				if(map1.containsKey(k)) {
					int t1 = map1.get(k);
					int t2 = map2.get(k);
					if(S.charAt(t1) != T.charAt(t2)) {
						System.out.println(-1);
						System.exit(0);
					}
				}
			}
			System.out.println(L);
		}else {
			System.out.println(L);
		}



	}
	static int gcd(int m, int n) {
	    if(m < n) return gcd(n, m);
	    if(n == 0) return m;
	    return gcd(n, m % n);
	}
	static int lcm(int m, int n) {
		return m * n / gcd(m, n);
	}
}