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;
long long n, k;
bool vis[100005];
vector<long long> v, vec, H;
vector<long long>::iterator it;
long long freq[100005];
long long dp[100005];
long long call(int pos) {
  if (pos >= H.size()) return 0;
  if (dp[pos] != -1) return dp[pos];
  long long a = H[pos] + call(pos + 2);
  long long b = call(pos + 1);
  return dp[pos] = max(a, b);
}
long long answer;
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  if (n == 1) {
    cout << 1;
    return 0;
  }
  v.resize(n);
  for (register int i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  vec.push_back(v[0]);
  int cnt = 1, pos = 0;
  for (register int i = 1; i <= n - 1; i++) {
    if (v[i] != v[i - 1]) {
      vec.push_back(v[i]);
      freq[pos++] = cnt;
      cnt = 1;
    } else {
      cnt++;
    }
  }
  freq[pos++] = cnt;
  if (k == 1) {
    cout << vec.size();
    return 0;
  }
  for (register int i = 0; i < pos; i++) {
    if (vis[i] == 0) {
      long long P = i, Px = i;
      do {
        P = Px;
        H.push_back(freq[P]);
        vis[P] = 1;
        it = lower_bound(vec.begin(), vec.end(), vec[P] * k);
        if (it == vec.end()) break;
        Px = it - vec.begin();
      } while (*it == vec[P] * k);
      answer += call(0);
      for (register int j = 0; j < H.size(); j++) dp[j] = -1;
      H.clear();
    }
  }
  cout << answer;
}
