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: Runtime Error)

#include <iostream>
#include <iomanip>
#include <string>
#include <algorithm>
#include <functional>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <deque>
#include <cmath>
#include <cstdio>

using namespace std;

typedef long long ll;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)

ll lcm(ll a, ll b)
{
    if (a < b) {
        ll t = a;
        a = b;
        b = t;
    }
    ll aa = a;
    ll bb = b;
    ll r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    ll gcd = b;
    a = aa;
    b = bb;
    return a * b / gcd;
}

int main(int argc, char **argv)
{
    ll N, M;
    string S, T;
    cin >> N >> M >> S >> T;

    ll L = lcm(N, M);
    char X[L];
    REP(i, L) X[i] = '0';

    // set S
    REP(i, N) {
        X[(L / N) * i] = S[i];
    }
    bool f = true;
    REP(i, M) {
        if (X[(L / M) * i] != '0' && X[(L / M) * i] != T[i]) {
            f = false;
            break;
        }
    }
    if (f) cout << L << endl;
    else cout << -1 << endl;

    return 0;
}