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: Accepted)

import java.util.*;

public class Main {

    public static void main(String args[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        String s = sc.next();
        String t = sc.next();
        sc.close();

        // 主処理
        long l = lcm(n, m);
        int lpn = (int) (l / n);
        int lpm = (int) (l / m);

        boolean judge = true;
        Map<Integer, String> map = new HashMap<>();

        for (int i = 0; i < n; i++) {
            String w = String.valueOf(s.charAt(i));
            int index = (i * lpn) + 1;
            if (!map.containsKey(index) || w.equals(map.get(index))) {
                map.put(index, w);
            } else {
                judge = false;
                break;
            }
        }

        for (int i = 0; i < m; i++) {
            String w = String.valueOf(t.charAt(i));
            int index = (i * lpm) + 1;
            if (!map.containsKey(index) || w.equals(map.get(index))) {
                map.put(index, w);
            } else {
                judge = false;
                break;
            }
        }

        long result = judge ? l : -1;

        // 出力
        System.out.println(result);
    }

    public static long lcm(long a, long b) {
        long gcd = gcd(a, b);
        return a / gcd * b;
    }

    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}