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;
template <typename T>
std::vector<std::vector<std::vector<T>>> make_3d_vector(long long z,
                                                        long long y,
                                                        long long x,
                                                        T value = T{}) {
  return std::vector<std::vector<std::vector<T>>>(
      z, std::vector<std::vector<T>>(y, std::vector<T>(x, value)));
}
template <typename T>
std::vector<std::vector<T>> make_2d_vector(long long z, long long y,
                                           T value = T{}) {
  return std::vector<std::vector<T>>(z, std::vector<T>(y, value));
}
template <typename T>
std::vector<T> make_1d_vector(long long z, T value = T{}) {
  return std::vector<T>(z, value);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n + 1);
  vector<long long> visited(n + 1, 0);
  map<long long, long long> mp;
  v[0] = -1;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  long long ans = 0;
  sort(v.begin(), v.end());
  for (long long i = 1; i <= n; i++) {
    mp[v[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    if (visited[i] == -1) {
      continue;
    }
    ans++;
    visited[mp[v[i] * k]] = -1;
  }
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
}
