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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<ll> vl;
typedef vector<vector<ll> > vvl;
#define REP(var, a, b) for (int var = (a); var < (b); var++)
#define rep(var, n) for (int var = 0; var < (n); ++var)
#define ALL(c) (c).begin(), (c).end()
#define rALL(c) (c).rbegin(), (c).rend()
ll MOD = 1000000007;
ll INF = 1e18;

ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
  ll g = gcd(a, b);
  return a * b / g;
}
int main() {
  //
  ll n, m;
  cin >> n >> m;
  string s, t;
  cin >> s >> t;
  assert(s.size() == n && t.size() == m);

  ll l = lcm(n, m);
  ll nn = l / n;
  ll mm = l / m;
  map<ll, char> mp;
  for (ll i = 0; i < n; i++) {
    mp[i * nn] = s[i];
    // x[i * nn] = s[i];
  }
  for (ll i = 0; i < m; i++) {
    if (mp.find(i * mm) == mp.end()) continue;
    if (mp[i * mm] != t[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << l << endl;
  return 0;
}