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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define zero_pad(num) setfill('0') << std::right << setw(num)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;

ll GCD(ll a, ll b) {
    return (b > 0) ? GCD(b, a % b) : a;
}

ll LCM(ll a, ll b) {
    return a / GCD(a, b) * b;
}

int main() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    if(n % m == 0){
        bool v = true;
        rep(i, m)if(s[i*(n/m)] != t[i])v = false;
        if(v)cout << n << endl;
        else cout << -1 << endl;
    }else if(m % n == 0){
        bool v = true;
        rep(i, n)if(s[i] != t[i*(m/n)])v = false;
        if(v)cout << m << endl;
        else cout << -1 << endl;
    }else{
        bool v = true;
        int g = GCD(n, m);
        rep(i, g)if(s[i*(n/g)] != t[i*(m/g)])v = false;
        if(v)cout << LCM(n, m) << endl;
        else cout << -1 << endl;
    }
    
}