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;
#define REP(i, n) for(int i = 0; i < (int)(n); i++)
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define _printV(vec) for(int i = 0; i < SZ(vec); i++){cout << vec[i];if (i==SZ(vec)-1){cout<<endl;}else{cout << " ";}}
#define printV(vec) cout<<#vec<<":";_printV(vec)
#define _printA(arr,n) for(int i = 0; i < n; i++){cout << arr[i];if (i==n-1){cout<<endl;}else{cout << " ";}}
#define printA(arr) cout<<#arr<<":";_printA(arr,sizeof(arr)/sizeof(arr[0]))
#define LLINF = (1LL<<62)
#define puts(var) cout<<#var<<":"<<var<<" "
typedef long long ll;

ll lcm( int m, int n )
{
	if ( ( 0 == m ) || ( 0 == n ) )
		return 0;
	return ((m / __gcd(m, n)) * n);
}
 
 
int main()
{
  int N,M;
  cin >> N >> M;
  string S;
  string T;
  cin >> S;
  cin >> T;
 
  ll L = lcm(N, M);
 
  int step = __gcd(N,M);
  // cout << step << endl;
  for(int i=0;i<step;i++){
    // printf("===== i %d\n",i);
    // cout << L/M <<endl;
    // cout <<"S:" << S[i*N/step] << endl;
    // cout << L/N <<endl;
    // cout <<"T:" << T[i*M/step] << endl;
    if(S[i*N/step] != T[i*M/step]){
      cout << -1 << endl;
      return 0;
    }
  }
  cout << (ll) L << endl;
  return 0;
}