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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
vector<int> s;
map<int, int> ma;
int main() {
  int i, j, k, n, ans = 0;
  cin >> n >> k;
  for (j = 0; j < n; j++) {
    int x;
    cin >> x;
    s.push_back(x);
  }
  j = n - 1;
  sort(s.begin(), s.end());
  while (j >= 0) {
    int p = s[j];
    if (ma[p] != 0) {
      j--;
      continue;
    }
    if (p % k == 0) {
      vector<int>::iterator it = lower_bound(s.begin(), s.end(), p / k);
      if (*it == p / k && ma[*it] != 1 && *it != p) {
        ans++;
        ma[p] = 1;
        ma[*it] = 1;
      }
    }
    j--;
  }
  cout << (int)s.size() - ans;
  return 0;
}
