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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
string x(1000005, '*');
int main() {
  int n, m;
  string s, t;
  cin >> n >> m;
  cin >> s;
  int l = __gcd(n, m);
  l = n / l * m;
  x[0] = s[0];
  for(int i = 1; i < n; ++i) {
    x[i * (l / n)] = s[i];
  }
  cin >> t;
  bool flag = true;
  if(x[0] != t[0]) flag = false;
  for(int i = 1; i < m && flag; ++i) {
    if(x[i * (l / m)] != '*' && x[i * (l / m)] != t[i]) flag = false;
    x[i * (l / m)] = t[i];
  }

  if(flag) cout << l << endl;
  else cout << -1 << endl;
  return 0;
}