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 {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();

		int M = sc.nextInt();

		String S = sc.next();

		String T = sc.next();

		char[] s = new char[S.length()];

		char[] t = new char[T.length()];

		int gcd = 0;

		boolean flag = false;

		int c = 0;




		for(int i=0;i<S.length();i++) {

			s[i] = S.charAt(i);

		}

		for(int i=0;i<T.length();i++) {

			t[i] = T.charAt(i);

		}

		if(s[0] != t[0] ) {

			System.out.println(-1);

		}else if(S.equals(T)) {

			System.out.println(S.length());

		}else {

			gcd = gcd(S.length(),T.length());

			int a = S.length() / gcd;

			int b = T.length() / gcd;

			c = a * b * gcd;

			int d = 0;

			int i = 1;

			int e = a * b;

			//int max = Math.max(S.length(), T.length());

			while((e/a)<t.length && (e/b)<s.length) {


				if(t[(e/a)] != s[(e/b)]) {


					flag = true;

					break;
				}

				i++;

				e = (a * b) * i;


			}

			if(flag == false) {

				System.out.println(c);

			}else {

				System.out.println(-1);
			}


		}




	}

	public static int gcd(int a, int b) {

		if(b==0) return a;

		return gcd(b,a%b);
	}

}