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>
#define rep(i,n)for(long long i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

ll gcd(ll x,ll y){
  if(x%y==0) return y;
  else gcd(y,x%y);
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll n, m;
  string s, t;
  cin >> n >> m >> s >> t;
  ll G = gcd(n,m);
  ll L = (n*m)/G;
  
  if(s[0]!=t[0]){
    cout << -1 << endl;
    return 0;
  }
  if(G==1){
    cout << L << endl;
    return 0;
  }
  if(n>=m){
    ll N = n/G;
    ll M = m/G;
    bool ok = true;
    for(ll i=1;i<G;i++){
      if(s[N*i]!=t[M*i]) ok = false;
    }
    
    if(ok){
      cout << L << endl;
    }
    else{
      cout << -1 << endl;
    }
  }
  else{
    ll N = n/G;
    ll M = m/G;
    
    bool ok = true;
    for(ll i=1;i<G;i++){
      if(s[N*i]!=t[M*i]) ok = false;
    }
    
    if(ok){
      cout << L << endl;
    }
    else{
      cout << -1 << endl;
    }
  }
  return 0;
  
}