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.*;
import java.util.*;
 
public class Main {
	static final Reader in = new Reader();
	
	public static void main(String[] args) {
		int n=in.nextInt(), m=in.nextInt();
		char[] s=in.next().toCharArray(), t=in.next().toCharArray();
		int g=gcd(n, m);
		for(int i1=0, i2=0; i1<s.length; i1+=n/g, i2+=m/g) {
			if(s[i1]!=t[i2]) {
				System.out.println(-1);
				return;
			}
		}
		System.out.println((long)n*m/g);
	}
	
	static int gcd(int a, int b) {
		while((a%=b)>0&&(b%=a)>0);
		return a^b;
	}
	
	static class Reader {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		String next() {
			while(st==null||!st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
}