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 <sstream>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <set>
#include <queue>
#include <map>
#include <cmath>
#include <limits>
#include <numeric>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<ld> vld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define INF ((int)1e9)
#define INFLL ((ll)1e18)
#define MOD (1000000007LL)

ll gcd(ll n, ll m) {
    if (m > n) {
        return gcd(m, n);
    }
    if (m == 0) {
        return n;
    }
    return gcd(m, n - n/m * m);
}

int main(int argc, const char * argv[]) {
    ios_base::sync_with_stdio(false);
    ll N, M;
    string S, T;
    cin >> N >> M >> S >> T;
    
    if (M > N) {
        swap(M, N);
        swap(S, T);
    }
    ll g = gcd(N, M);
    ll L = N*M/g;
    
    if (S[0] != T[0]) {
        cout << -1 << endl;
        return 0;
    }
    
    if (L == N) {
        for (ll m = 0; m < M; ++m) {
            if (S[N/M * m] != T[m]) {
                cout << -1 << endl;
                return 0;
            }
        }
    }
    
    cout << L << endl;

    return 0;
}