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 <vector>
#include <numeric>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define INF 1145141919
typedef long long int ll;
using namespace std;

template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
    for (T &x : vec)
        is >> x;
    return is;
}

//*********************************************************
// 最大公約数（Greatest Common Divisor）を返す。
// 引数に０がある場合は０を返す。
//*********************************************************
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（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)
}//lcm

int main(){
    ll n,m;
    cin >> n >> m;
    string s,t;
    cin >> s >> t;

    ll res = lcm(n,m);
    //cout << res << endl;
    ll LN = res / n;
    ll LM = res / m;
    ll cnt = max(LN,LM);

    REP(i,res/(LN*LM)){
        if(s[i*LM]!=t[i*LN]){
            cout << "-1" << endl;
            return 0;
        }
    }
    cout << res << endl;

    return 0;
}