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<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#include<cmath>
#include<algorithm>
#include<bitset>
#include<map>
#include<functional>
#include<set>
#include<numeric>

#pragma region
using namespace std;
#define FOR(i,r,n) for(ll i = (ll)(r); i < (ll)(n); i++)
#define rep(i,n) FOR(i,0LL,n)
#define RFOR(i,r,n) for(ll i=(ll)(n-1);i>=r;i--)
#define rrep(i,n) RFOR(i,0LL,n)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define COUNT(a,y,x) upper_bound(all(a), y) - lower_bound(all(a), x)
#define UNIQUE(a) sort(all(a)); a.erase(unique(all(a)), a.end())
#define SUM(a) accumulate(all(a),0LL)
#define scan(x) scanf("%lld",&x)
#define print(x) printf("%lld\n",x)
#define print2(x) printf("%lld ",x)
#define pb push_back
typedef long long int ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef map<ll, ll> MAP;
const ll inf = 2222222222222222222LL;
const ll mod = 1000000007LL;

ll n = 0, m = 0, ans = 0, cnt = 0, tmp = 0, ma = -inf, mi = inf;
string s;
bool ok = true, flag = false;
ll dx[5] = { 1,-1,0,0,0 }, dy[5] = { 0,0,1,-1,0 };
ll ddx[9] = { 1,-1,0,0,1,1,-1,-1,0 }, ddy[9] = { 0,0,1,-1,1,-1,1,-1,0 };
#pragma endregion
#define MAX 222222

ll gcd(ll x, ll y) {
  if (x < y) swap(x, y);
  return y ? gcd(y, x%y) : x;
}

int main(void) {
  string t;
  cin >> n >> m;
  cin >> s;
  cin >> t;
  ok = true;
  ll g = gcd(n, m);
  ll x = n / g, y = m / g;
  rep(i, g) {
	if (s[i*x] != t[i*y]) ok = false;
  }
  if (ok) cout << n * y << endl;
  else cout << "-1" << endl;


  return 0;
}