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 <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <tuple>
#include <vector>

using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define reps(i, f, n) for (int i = (f); i < (n); i++)
#define repv(v) for (auto it = (v).begin(); it != (v).end(); it++)
ll const INF = 1LL << 60;

int main() {
    ll N, M;
    cin >> N >> M;
    string S, T;
    cin >> S >> T;

    ll ans = -1;

    ll L = N / __gcd(N, M) * M;
    ll a = L / N;
    ll b = L / M;
    ll c = a / __gcd(a, b) * b;
    ll d = c / a;
    ll e = c / b;

    bool flg = true;
    for (ll k = 0; k < min(N / d, M / e); k++) {
        if (S[k * d] != T[k * e]) {
            flg = false;
            break;
        }
    }

    if (flg)
        cout << L << endl;
    else
        cout << "-1" << endl;

    return 0;
}