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: Runtime Error)

import java.util.*;

public class Main {
	public static void main(String args[]) {
		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();
		
		int num = (int)lcm(N, M);
		
		boolean flag = true;
		int cnt = 0;
		while(flag) {
			flag = false;
			cnt++;
			num *= cnt;
			if(cnt > 5) {
				num = -1;
				break;
			}
			String str[] = new String[10000000];
			Arrays.fill(str, "#");
			int n_num = num / N;
			int m_num = num / M;
			for(int i = 0; i < N; i++) {
				str[i * n_num + 1] = S.substring(i, i + 1);
			}
			for(int i = 0; i < M; i++) {
				if(!str[i * m_num + 1].equals("#") && !str[i * m_num + 1].equals(T.substring(i, i + 1))) {
					flag = true;
				}
			}
		}
		System.out.println(num);
	}
	
	private static long lcm(long m, long n) {
	    return m * n / gcd(m, n);
	}
	
	private static long gcd(long m, long n) {
	    if(m < n) return gcd(n, m);
	    if(n == 0) return m;
	    return gcd(n, m % n);
	}
}