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>
#include <iostream>
#include <string>
using namespace std;
//#define sz(x) int((x).size())

/*int gcd(int a, int b) {
  if(a < b) gcd(b, a);
  int r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
  }*/


int main() {
  int N,M;
  cin >>N >> M;
  string S,T;
  cin >> S >> T;
  long long int GCD = __gcd(M,N);
  int n = N/GCD;
  int m = M/GCD;
  bool out=false;
  for (int i=0; i<GCD; i++){
    //      if(S[i*n] != T[i*m]){
    if(S[i*(N/GCD)] != T[i*(M/GCD)]){  
      //        cout <<  -1 << endl;
      //        return 0;
      out = true;
      }
      
  }
  if(out)
    cout<<-1<<endl;
  else
    cout<<M*(N/GCD)<<endl;
  //  cout << M*N/GCD << endl;
  //  return 0;
}