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 <list>
#include <algorithm>
#include <string>
#include <sstream>
#include <stack>
#include <iomanip>
#include <numeric>
#include <queue>
#include <climits>
#include <set>
#include <complex>
#include <cmath>
#include <cstring>
#include <map>
using namespace std;
using ll = long long;
#define MOD 1000000007
#define INF 1LL << 59
using ld = long double;

const int MAX = 510000;

ll gcd(ll a, ll b)
{
    if (a < b)
        swap(a, b);
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    //cout << fixed << setprecision(5);

    ll N, M;
    cin >> N >> M;

    string s, t;
    cin >> s >> t;

    ll d = N * M / gcd(N, M);

    ll t1 = d / M;
    ll t2 = d / N;

    ll T = t1 * t2;
    ll T1 = T;
    for (int i = 0;; ++i)
    {
        if (T / t1 >= M || T / t2 >= N)
            break;
        if (t[T / t1] != s[T / t2])
        {
            cout << -1 << endl;
            return 0;
        }
        T += T1;
    }
    cout << d << endl;

    return 0;
}