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;

unsigned int gcd(unsigned int u, unsigned int v) {
    while (v != 0) {
        unsigned int r = u % v;
        u = v;
        v = r;
    }
    return u;
}

unsigned int lcm (unsigned int a, unsigned int b)
{
    if ((a == 0) || (b == 0)) return 0;
    return a * b / gcd(a, b);
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M;
    string S, T;

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

    int LCM = lcm(N, M);

    int S_step = LCM / M;
    int T_step = LCM / N;

    int n = 0;
    int m = 0;

    while (n < N && m < M) {
        if (S[n] != T[m]) {
            cout << -1 << endl;
            return 0;
        }

        n += S_step;
        m += T_step;
    }

    cout << LCM << endl;
    return 0;
}