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;

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
long long 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() {
    string s, t;
    int n, m;
    cin >> n >> m >> s >> t;
    int GCD = gcd(n, m);
    long long ans = lcm(n, m);
    int n2 = n / GCD, m2 = m / GCD;
    for (int i = 0; i < GCD; i++) {
        if (s[i * n2] != t[i * m2]) {
            ans = -1;
            break;
        }
    }
    cout << ans;
}