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.*;

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();
        char[] sa = s.toCharArray();
        String t = sc.next();
        char[] ta = t.toCharArray();

        int l = lcm(n, m);
        
        char[] x = new char[100000];

        for (int i = 0; i < sa.length; i++) {
            if (i == 0) {
                x[i] = sa[i];
            } else {
                int j = i * (l/sa.length) + 1;
                x[j] = sa[i];
            }
        }

        int rel = l;
        for (int i = 0; i < ta.length; i++) {
            if (i == 0) {
                if (x[i] != ta[i]) {
                    rel = -1;
                    break;
                } 
            } else {
                int j = i * (l/ta.length) + 1;
                char tt = x[j];
                if (tt != '\u0000' && tt != ta[i]) {
                    rel = -1;
                    break;
                }
            }
        }

        if (rel != -1) {
            
        }

        System.out.println(rel);
        sc.close();
    }

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