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>
const long double pie = 3.1415926535;
using namespace std;
long long fpow(long long n, long long p) {
  long long m = 1;
  while (p) {
    if (p % 2) m *= n;
    p >>= 1;
    n *= n;
  }
  return m;
}
long long mfpow(long long n, long long p, long long M) {
  long long m = 1;
  n %= M;
  while (p) {
    if (p % 2) m = (m * n) % M;
    n = (n * n) % M;
    p >>= 1;
  }
  return m % M;
}
const vector<long long> days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool br1, br2, br;
char ch, ch1, ch2;
long long n, k, x, y, z, mini, maxi, l, ind, ini, sum, t, len, r, w, imini, m;
string s, s1, s2;
map<long long, bool> mp1;
long long a[(int)1e6 + 100];
void solve() {
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (long long i = 0; i < n; i++) {
    if (a[i] % k || mp1[a[i] / k] == 0) {
      m++;
      mp1[a[i]] = 1;
    }
  }
  cout << m << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}
