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.Scanner;

public class Main {

	@SuppressWarnings("resource")
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int n=scanner.nextInt();
		int m=scanner.nextInt();
		String s=scanner.next();
		String t=scanner.next();
		long l=lcm(n,m);
		String x[]=new String[(int)l+1];
		for(int i=0;i<l+1;i++) {
			x[i]="NO";
		}
		for(int i=0;i<n;i++) {
			x[(int) (i*l/n+1)]=String.valueOf(s.charAt(i));
		}
		for(int i=0;i<m;i++) {
			int ind=(int) (i*l/m+1);
			String tmp=String.valueOf(t.charAt(i));
			if(tmp.equals(x[ind])) {
				continue;
			}else if(x[ind].equals("NO")) {
				x[ind]=tmp;
			}
			else {
				System.out.println(-1);
				return;
			}
		}
		System.out.println(l);
	}
	
	/*******************************************************************
	最大公約数
	*******************************************************************/

	static long gcd(long a, long b) {
			long temp;
			while ((temp = a % b) != 0) {
				a = b;
				b = temp;
			}
			return b;
		}


	/*******************************************************************
	最小公倍数
	*******************************************************************/

		 static long lcm(long m, long n) {
			return m / gcd(m, n) * n;
		}

}