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 int power(long long int x, long long int y, long long int z) {
  if (y == 0)
    return 1;
  else {
    long long int p = power(x, y / 2, z);
    if (y % 2 == 0)
      return (p * p) % z;
    else
      return (x * ((p * p) % z)) % z;
  }
}
long long int gcd(long long int x, long long int y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long int po(long long int x) {
  long long int c = 0;
  while (x > 0) {
    x = x & (x - 1);
    c++;
  }
  return c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  long long int n, k;
  cin >> n >> k;
  long long int a[n], ans = 0;
  map<long long int, long long int> mp;
  for (int i = 0; i < n; i++) cin >> a[i];
  if (k == 1) {
    cout << n;
    exit(0);
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) mp[a[i]] = i + 1;
  for (int i = n - 1; i >= 0; i--) {
    long long int x = a[i], c = 0, y = x;
    while (x % k == 0 && mp[x / k] > 0 && x > 0) {
      c++;
      x = x / k;
    }
    if (c > 0) {
      for (int j = 1; j <= c; j = j + 2) {
        a[mp[y / pow(k, j)] - 1] = -1;
      }
    }
  }
  long long int d = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] > 0) d++;
  }
  cout << d;
}
