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 {
    public static void main(String[] args) {
        // 自分の得意な言語で
        // Let's チャレンジ！！

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
      int m=sc.nextInt();
      String x=sc.nextLine();
      x=sc.nextLine();
      String y=sc.nextLine();
      //System.out.println(x+" "+y+" "+x.split("",0)[0].equals(y.split("",0)[0]));
      if(n%m==0||m%n==0||!x.split("",0)[0].equals(y.split("",0)[0])){
      System.out.println(-1);
      }
      else {
      System.out.println(lcm(m, n));
      }
        
    }
   static int gcd(int m, int n) {
        if(m < n) return gcd(n, m);
        if(n == 0) return m;
        return gcd(n, m % n);
    }
    
    static int lcm(int m, int n) {
        return m * n / gcd(m, n);
    }
}