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++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ll long long

/* aとbの最大公約数を求める. */
ll gcd(ll a, ll b) {
	if (b > a) return gcd(b, a);
	if (b == 0) return a;
	return gcd(b, a % b);
}

/* aとbの最小公倍数を求める. */
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return (a / g) * b;
}

int main() {
	int N, M;
	cin >> N >> M;
	ll L = lcm(N, M);
	ll G = gcd(N, M);
	string S, T;
	cin >> S >> T;
	int m = (int)M / G;
	int n = (int)N / G;
	for (int j = 0; j < M; j += m) {
		int i = j / m * n;
		if (S[i] != T[j]) {
			cout << -1 << endl;
			return 0;
		}
	}
	cout << L << endl;
	return 0;
}