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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long

int gcd(int a, int b){
	if(b == 0){
		return a;
	}
	else{
		return gcd(b, a % b);
	}
}

signed main(){
	int N, M, i, j, a, b, g;
	scanf("%lld%lld", &N, &M);
	char *S = (char *)malloc(sizeof(char) * (N + 1));
	char *T = (char *)malloc(sizeof(char) * (M + 1));
	scanf("%s", S);
	scanf("%s", T);
	g = gcd(N, M);
	a = N / g;
	b = M / g;
	for(i = 0, j = 0; i * a < N && j * b < M; i++, j++){
		if(S[i * a] != T[j * b]){
			printf("-1\n");
			return 0;
		}
	}
	printf("%lld\n", (N * M) / g);
	return 0;
}