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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
public class Main {
    public static void main(String[] args) throws IOException {
	Main m = new Main();
	m.solve();
    }

    private void solve() throws IOException {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	int[] nm = Arrays.stream(br.readLine().split("[\\s]+"))
	    .mapToInt(Integer::parseInt).toArray();
	int n = nm[0];
    	int m = nm[1];
	String S = br.readLine();
	String T = br.readLine();
	br.close();
	int gcd = n > m ? gcd(n,m) : gcd(m,n);
	long lcm = (long)n*(long)m/(long)gcd;
	for (int i=0;i<n;i++) {
	    if (((long)i*(long)m)%(long)n != 0) continue;
	    long j = ((long)i*(long)m)/(long)n;
	    if ((int)j == m) continue;
	    char s = S.charAt(i);
	    char t = T.charAt((int)j);
	    if (s != t) {
		System.out.println(-1);
		System.exit(0);
	    }
	}
	System.out.println(lcm);
    }

    private int gcd(int n,int m) {
	if (n%m == 0) return m;
	return gcd(m,n-m*(n/m));
    }
}