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 <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)

ll lcm(ll a, ll b){
  ll tmp, r, x = a * b;
  if(a < b){
    tmp = a;
    a = b;
    b = tmp;
  }
  r = a % b;
  while(r != 0){
    a = b;
    b = r;
    r = a % b;
  }
  return x / b;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

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

  ll lc = lcm(n, m);

  if(lc == n * m){
    if(s[0] == t[0]) cout << lc << endl;
    else cout << -1 << endl;
    return 0;
  }

  ll gc = n * m / lc;
  ll rec = 0;
  REP(i, gc){
    if(s[i * (n / gc)] != t[i * (m / gc)]) rec++;
  }
  if(rec) cout << -1 << endl;
  else cout << lc << endl;
}