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.*; 
class Main
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        String s = sc.next(), t = sc.next();

        long l = calc(n, m);
        int x = (int)l / m;
        int y = (int)l / n;

        boolean bl = true;
        for(int i = 0; i < s.length(); i += x){
            if(s.charAt(i) != t.charAt((int)(i / x * y))){
                System.out.println(-1);
                bl = false;
                break;
            }
        } 
        if(bl)
            System.out.println(l);

    }
    static long calc (int n, int m)
    {
        int x = Math.min(n, m);
        int y = Math.max(n, m);
        for(int i = 1; i < x;i++)
        {
            if(y * i % x == 0){
                return y * i;
            }
        }
        return y * x;
    }
 }