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: Wrong Answer)

#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const string ln = "\n";
constexpr int INF = 1001001001;
constexpr int MOD = 1000000007;

// lcm
ll lcm(ll x, ll y) { return (x * y) / __gcd(x, y); }

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);

  ll n, m;
  cin >> n >> m;
  string s, t;
  cin >> s >> t;

  ll l = lcm(n, m);

  bool ok = true;
  if (n % m == 0 || m % n == 0) {
    if (n > m) {
      for (int i = 0; i < m; i++) {
        if (t[i] != s[i * l / m]) {
          ok = false;
          break;
        }
      }
    } else {
      for (int i = 0; i < n; i++) {
        if (s[i] != t[i * l / n]) {
          ok = false;
          break;
        }
      }
    }
  }

  cout << (ok ? l : -1) << ln;

  return 0;
}