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: Wrong Answer)

/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 1e5 + 10;

int main() {
	fast_cin();
	int n, m;
	string s, t;
	cin >> n >> m >> s >> t;
	int l = (n * ll(m)) / __gcd(n, m);
	for (int i = 0; i < n; ++i) {
		if (i * m % n == 0) {
			ll j = (i * ll(m)) / n;
			if (j >= 0 and j < m and s[i] != t[j]) {
				cout << "-1";
				return 0;
			}
		}
	}
	cout << l << endl;
}