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>
int gcd(int a, int b)
{
	int r = a % b;
	while (r > 0)
	{
		a = b;
		b = r;
		r = a % b;
	}
	return b;
}
int main()
{
	int n, m;
	scanf("%d %d", &n, &m);
	char s[100005], t[100005];
	scanf("%s%s", s, t);
	int g;
	g = gcd(n, m);
	int i;
	for (i = 0; i < g; i++)
	{
		if (s[i * n / g] != t[i * m / g])
		{
			printf("-1\n");
			return 0;
		}
	}
	printf("%lld\n", (long long int)n * (long long int)m / (long long int)g);
	return 0;
}