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.*;

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/5 22:51
 */
public class Main {

    final static int MOD = 1000000007;

    private static int comb(int n, int m) {
        return 1;
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    @SuppressWarnings("unchecked")
    private static void solve(Scanner sc) {
        int n = sc.nextInt();
        int m = sc.nextInt();
        char s[] = sc.next().toCharArray();
        char t[] = sc.next().toCharArray();
        if (s[0] != t[0]) {
            System.out.println(-1);
            return;
        }
        int gcd = gcd(n, m);
        long lcm = n * m / gcd;
        int t1 = n / gcd;
        int t2 = m / gcd;
        while (t1 < n && t2 < m) {
            if (s[t1] != t[t2]) {
                System.out.println(-1);
                return;
            }
            t1 <<= 1;
            t2 <<= 1;
        }
        System.out.println(lcm);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    public static void main(String args[]) {

        solve(new Scanner(System.in));
    }
}