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: Runtime Error)

import java.util.Deque;
import java.util.ArrayDeque;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;


public class Main {
    static long lcm (long a, long b) {
    	long temp;
    	long c = a;
    	c *= b;
    	while((temp = a%b)!=0) {
    		a = b;
    		b = temp;
    	}
    	return (long)(c/b);
    }

    
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] nm = br.readLine().split(" ");
        int n = Integer.parseInt(nm[0]);
        int m = Integer.parseInt(nm[1]);
        String s = br.readLine();
        String t = br.readLine();
        
        long x = lcm(n, m);
        long ans = x;
        //outer: for (int i = x; (i <= (n-1)*i/n + 1) || (i <= (m-1)*i/m + 1); i += x) {
        //outer: for (long i = x; i <= x; i += x) {
            //System.out.println(i);
        for (int j = 0; j < n; j += x/m) {
            //for (int k = 0; k < n; k += 1) {
                //if (j*i/n+1 == k*i/m+1) {
                //if (j*l/n == ) {
                    //System.out.println("aaa");
                    if (s.charAt(j) != t.charAt(j*m/n)) {
                        ans = -1;
                        break;
                    }
                //}
                
            //}
        }
        //}
        System.out.println(ans);
    }
}