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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, a, b) for(int i = a; i < b; i++)
#define  all(x) (x).begin(),(x).end()     // 昇順ソート
#define  rall(v) (v).rbegin(), (v).rend() // 降順ソート
#define  FastIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
using P = pair<int,int>;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return true; } return false; }

int main(){
    ll n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    ll lcm = n / __gcd(n, m) * m;
    map<ll, char> mp;
    for(int i = 0; i < s.size(); i++){
        ll j = i * (lcm/n) + 1;
        mp[j] = s[i];
    }
    for(int i = 0; i < t.size(); i++){
        ll j = i * (lcm/m) + 1;
        if (mp.find(j) != mp.end()){
            if (mp[j] != t[i]){
                cout << -1 << endl;
                return 0;
            }
        }
    }
    cout << lcm << endl;
    return 0;
}