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.*;

class Main{
	
	public static void main(String args[]){

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();

		String A = sc.next();
		String B = sc.next();
		boolean flag = true;

		if(N==M){
			if(A.equals(B)) System.out.println(A);
			else System.out.println("-1");
			flag = false;
		}else{
			if(N < M){
				int t = N;
				N = M;
				M = t;
				String st = A;
				A = B;
				B = st;
			}
		}

		if(flag){
			int tmp;
			int x=N;
			int y=M;
        	while ((tmp = x % y) != 0) {
            	x = y;
            	y = tmp;
        	}
        	int gcd = y;
        	int lcm = M*N/gcd;
        	boolean f = true;
        	for(int n=0,k=0;n<lcm;n+=(M/gcd)*(N/gcd),k++){
        		if(A.charAt(k*N/gcd) != B.charAt(k*M/gcd)){
        			f = false;
        		}
        	}

        	if(f) System.out.println(lcm);
        	else System.out.println("-1");


		}
	}
}