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>
using namespace std;
typedef long long ll;

ll gcd(ll a, ll b){
  if(a%b==0LL) return b;
  return gcd(b, a%b);
}

int main(){
  ll n, m; cin >> n >> m;
  string s, t;  cin>> s >> t;
  ll L = n*m/gcd(n, m);
  ll ans=L;

  if(s!=t && n==m) ans = -1;
  else{
    ll divn=L/n, divm=L/m, ic=0, tc=0;
    for(ll i=0; i<L; i++){
      if(i%divn==0 && i%divm==0){
        if(t[ic]!=t[tc]) ans = -1;
        ic += 1; tc += 1; 
      }else if(i%divn==0) ic+=1;
      else if(i&divm==0) tc += 1;
    }
 
  }

    cout << ans << endl;

  
  
}