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 rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMSET(v, h) memset((v), h, sizeof(v))
#define pb push_back
#define mp make_pair
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
typedef long long ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
using namespace std;

int main(){
  ll N,M;
  cin >> N >> M;
  string S,T;
  cin >> S >> T;
  if(S[0]!=T[0]){
    cout << -1 << endl;
    return 0;
  }
  if(N==M){
    if(S!=T){
      cout << -1 << endl;
      return 0;
    }
    else{
      cout << N << endl;
      return 0;
    }
  }
  ll mx = 0;
  ll mn = 0;
  string& mx_s = S;
  string& mn_s = T;
  if(N>M){
    mx = N;
    mn = M;
    string& mx_s = S;
    string& mn_s = T;
  }
  else{
    mx = M;
    mn = N;
    string& mx_s = T;
    string& mn_s = S;
  }
  ll MN = M*N;
  for(ll i=1;i<=mn-1;++i){
    ll div = (i*MN/mn) / (MN/mx);
    if( (i*MN/mn) % (MN/mx) == 0 && div <= (mx-1)){
      
      if(mn_s[i]!=mx_s[div]){
          cout << -1 << endl;
          return 0;
      }
      
    }
  }
  for(ll i=1;i<=mx-1;++i){
    ll div = (i*MN/mx) / (MN/mn);
    if( (i*MN/mx) % (MN/mn) == 0 && div <= (mn-1)){
      
      if(mx_s[i]!=mn_s[div]){
          cout << -1 << endl;
          return 0;
      }
      
    }
  }
  

  cout << (MN / __gcd(N,M)) << endl;


  return 0;
}