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: Accepted)

#include <bits/stdc++.h>
const char nl = '\n';
using namespace std;
typedef long long ll;
typedef long double ld;
typedef complex<ld> pt;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m; cin >> n >> m;
    string s, t; cin >> s >> t;
    ll len = (ll) n * m / __gcd(n, m);
    unordered_map<ll, int> dumb;
    for (int i = 0; i < n; i++) {
        dumb[len/n*i] = i;
    }
    for (int i = 0; i < m; i++) {
        if (dumb.count(len/m*i) && s[dumb[len/m*i]] != t[i]) {
            puts("-1");
            return 0;
        }
    }
    cout << len << nl;
    return 0;
}