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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const ll ZERO = 0;
const ll INF = 1e14;
const ll MOD = 1000000007;
const double PI = 3.1415926535897;

ll gcd(ll a,ll b) {
	if(a == 0 || b == 0) {
		return max(a,b);
	}
	else {
		if(a < b) {
			ll tmp = a;
			a = b;
			b = tmp;
		}
		ll r = a % b;
		while(r != 0) {
			a = b;
			b = r;
			r = a % b;
		}
		return b;
	}
}

int main() {
    ll N,M;
    cin >> N >> M;
    string S,T;
    cin >> S >> T;
    ll G = gcd(N,M);
    ll n = N / G;
    ll m = M / G;
    bool ret = true;
    for(int i = 0;i < N;i++) {
        if(i * n >= N) {
            break;
        }
        else if(S.at(i * n) != T.at(i * m)) {
            ret = false;
        }
    }
    if(ret == true) {
        cout << N * M / G << endl;
    }
    else {
        cout << -1 << endl;
    }
}