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.Scanner;

public class Main {

	public static void main(String args[]) {
		    Scanner sc = new Scanner(System.in);

		    int n = sc.nextInt();
		    int m = sc.nextInt();
		    String s = sc.next();
		    String t = sc.next();
		    
		    int gcd = gcd(n,m);
		    int count = 0;
		    for(int i = 0; i < gcd; i++) {
		    	if(s.substring(0+i*n/gcd,1+i*n/gcd).equals(t.substring(0+i*m/gcd, 1+i*m/gcd))) {
		    		count++;
		    	}
		    }
		    if(count == gcd) {
		    	System.out.println(n*m/gcd);
		    }else {
		    	System.out.println(-1);
		    }
		    
		    
		    sc.close(); }
	
	public static  int gcd(int a, int b) {
	    if (a < b) {
	        int tmp = a;
	        a = b;
	        b = tmp;
	    }

	    int r = -1;
	    while (r != 0) {
	        r = a % b;
	        a = b;
	        b = r;
	    }

	    return a;    
	}

}