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 <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#define ll int
using namespace std;
ll gcd(int a,int b);


int main(){
  int n,m;
  scanf ("%d %d",&n,&m);
  string s,t;
  cin >> s;
  cin >> t;
  ll g=gcd(s.size(),t.size());
  ll sa=s.size()/g;
  ll ta=t.size()/g;
  ll i;
  for(i=0;i<g;i++){
    if(s[sa*i]!=t[ta*i]) break;
  }
  if(i==g){
    cout << g*sa*ta << endl;
  }
  else{
    cout << "-1" << endl;
  }
}


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