A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.

You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

All the numbers in the lines are separated by single spaces.

Output

On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.

Examples

Input

6 2
2 3 6 5 4 10


Output

3

Note

In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
using grafo = vector<vi>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
template <class T>
using PR = pair<T, T>;
const ll MOD = 1e9 + 7;
const int MX = 1000100;
const ll INF = 1e18;
const ll INFI = 1e9;
const db PI = acos((db)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
void solve() {
  ll n, k;
  cin >> n >> k;
  vl a(n);
  map<ll, ll> m;
  for (int i = (0); i < (n); ++i) {
    cin >> a[i];
    m[a[i]] = 1;
  }
  for (int i = (0); i < (n); ++i) {
    if (m[a[i] * k]) m[a[i] * k] = a[i];
  }
  sort(begin(a), end(a));
  ll c = 0;
  map<ll, ll> m1;
  for (int i = (0); i < (n); ++i) {
    if (m1[a[i]]) continue;
    m1[a[i] * k] = 1;
    c++;
  }
  cout << c << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
