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: Time Limit Exceeded)

/**
 * Created at 10:25 on 2019/03/05
 */

import java.util.*;

public class Main {

  static Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {

    int N = sc.nextInt();
    int M = sc.nextInt();
    String S = sc.next();
    String T = sc.next();

    long L = lcm(N, M);
    int g = gcd(N, M);
    int intervalS = (int)(L / M);
    int intervalT = (int)(L / N);

    for (int i=0; i<g; i++) {
      if (S.charAt(intervalS*i) != T.charAt(intervalT*i)) {
        System.out.println(-1);
        return;
      }
    }

    System.out.println(L);

  }

  static long lcm(long a, long b) {
    long ans = 1;
    for (long i=2; i<=a*b; i++) {
      if (i % a == 0 && i % b == 0) {
        ans = i;
        break;
      }
    }
    return ans;
  }

  static int gcd(int a, int b) {
    int ans = 1;
    int min = Math.min(a, b);
    for (int i=2; i<=min; i++) {
      if (a % i == 0 && b % i == 0) {
        ans = i;
      }
    }
    return ans;
  }

}