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.*;

public class Main {

    long INF = Long.MAX_VALUE;

    public static void main(String[] args) {
        new Main().solve();
    }

    void solve() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        long ans = (N / gcd(N, M))*M;
        String S = sc.next();
        String T = sc.next();
        char[] s = S.toCharArray();
        char[] t = T.toCharArray();
        long g = gcd(N, M);
        if (g == 1) {
            g = ans;
        }
        int c1 = 1;
        int c2 = 1;
        int a = 1;
        int b = 1;
        if (s[0] != t[0]) {
            System.out.println("-1");
        }
        lavel:
        for (int i = 1; i < N; i++) {
            if ((i * (ans / N) + 1) % g == 0) {
                for (int j = c2; j < M; j++) {
                    if ((j * (ans / M) + 1) % g == 0) {
                        c2 = j + 1;
                        if (s[a] != t[b]) {
                            System.out.println("-1");
                            return;
                        }
                        b++;
                        continue lavel;
                    }
                }
            }
            a++;
        }
        System.out.println(ans);
    }

    long gcd(int m, int n) {
        if (m < n) {
            return gcd(n, m);
        }
        if (n == 0) {
            return m;
        }
        return gcd(n, m % n);
    }
}