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;
long long gcd(long long a, long long b){
  while(a > 0 && b > 0){
    if(a > b){
      a %= b;
    } else {
      b %= a;
    }
  }
  return a + b;
}
int main(){
    long long n, m;
    cin >> n >> m;
    long long g = n * m / gcd(n, m);
    map<long long, char> fu;
    string s, t;
    cin >> s >> t;
    for(int i = 0; i < n; i++){
      fu[i * g / n] = s[i];
    }
    for(int i = 0; i < m; i++){
      if(fu.find(i * (g / m)) != fu.end()){
        if(fu[i * (g / m)] != t[i]) return !printf("-1\n");
      } else {
        fu[i * (g / m)] = t[i];
      }
    }
    cout << g << endl;
    return 0;
}