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<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<string>
#include<utility>
#include<vector>

#define rep(i, a, n)  for(int i=a; i<n; i++)
#define per(i, a, n)  for(int i=n-1; i>=a; i--)
#define all(x)  (x).begin(), (x).end()
using namespace std;

int main(){
  int n, m;
  string s, t;
  cin >> n >> m >> s >>t;
  int ans;
  if(s[0] != t[0]){
    ans = -1;
  }
  if(ans != -1){
    rep(l, min(n, m), 100000){
      if(ans>0) break;
      if(l%n != 0 || l%m !=0) continue;
      string tmp(l, 'A');
      int divN = l/n, divM = l/m;
      ans = 0;
      rep(i, 0, n){
        tmp[i * divN] = s[i];
      }
      rep(i, 0, m){
        if(tmp[i * divM] != 'A' && tmp[i * divM] != t[i]){
          ans = -1;
          break;
        }
        else{
          tmp[i * divM] = t[i];
        }
      }
      if(ans == 0){
        ans = l;
        break;
      }
    }
  }
  cout << ans << endl;
  return 0;
}