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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define fi first
#define se second
typedef pair<ll, ll> P;
using VP = vector<P>;
using VVP = vector<VP>;
using VI = vector<ll>;
using VVI = vector<VI>;
using VVVI = vector<VVI>;
const int inf = 1e9 + 7;
const ll INF = 1LL << 61;
const ll mod = 1e9 + 7;

template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int i, j;
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    int g = gcd(n, m);
    for (i = 0; i < n; i++) {
        if (i * m % n == 0) {
            if (s[i] != t[i * m / n]) {
                cout << -1 << endl;
                return 0;
            }
        }
    }
    cout << n / g * m << endl;
}