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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define vout(x) rep(i,x.size()) cout << x[i] << " "
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9;
const ll inf_l = 1e18;
const int MAX = 1e5;

//最大公約数と最小公倍数
ll gcd(ll a, ll b) {return(b?gcd(b,a%b):a);}
ll lcm(ll a, ll b) {return(a/gcd(a,b)*b);}

int main() {
    int n, m; cin >> n >> m;
    string s, t; cin >> s >> t;
    ll g = lcm(n,m);
    ll h = lcm(g/n,g/m);
    bool ok = true;
    for (int i = 0; i <= (n-1)*(g/n); i += h) {
        if (i <= (m-1)*(g/m)) {
            if (s[i/(g/n)] != t[i/(g/m)]) ok = false;
        }
    }
    if (ok) cout << g << endl;
    else cout << -1 << endl;
}