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 <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>

typedef long long ll;
using namespace std;

int gcd( int m, int n )
{
	// 引数に０がある場合は０を返す
	if ( ( 0 == m ) || ( 0 == n ) )
		return 0;
	
	// ユークリッドの方法
	while( m != n )
	{
		if ( m > n ) m = m - n;
		else         n = n - m;
	}
	return m;
}//gcd
//最小公倍数を求める。
//*********************************************************
// 最小公倍数（Least Common Multiple）を返す。
// 引数に０がある場合は０を返す。
//*********************************************************
int lcm( int m, int n )
{
	// 引数に０がある場合は０を返す
	if ( ( 0 == m ) || ( 0 == n ) )
		return 0;
	
	return ((m / gcd(m, n)) * n); // lcm = m * n / gcd(m,n)
}//

int main()
{
    int N, M, tmp, a, b, k, l;
    string S, T;

    cin >> N >> M;
    cin >> S >> T;

    tmp = lcm(N, M);
    a = tmp/N;
    b = tmp/M;

    if(N >= M) {
        k = N;
        l = M;
    } else {
        k = M;
        l = N;
    }


    for(int i=0; i<k; i++){
        for(int j=i; j<l; j++){
            if((i*a == j*b) && (S[i] != T[j])){
                cout << "-1" << endl;
                return 0;
            }
            if(i*a > j*b){
                break;
            }
        }
    }

    cout << tmp << endl;
    return 0;
}