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 <class T, class T1>
int chkmin(T &x, const T1 &y) {
  return x > y ? x = y, 1 : 0;
}
template <class T, class T1>
int chkmax(T &x, const T1 &y) {
  return x < y ? x = y, 1 : 0;
}
long long MAXN = 9223372036854775807, mod = 998244353;
bool comp(pair<long long, pair<long long, long long>> &a,
          pair<long long, pair<long long, long long>> &b) {
  if (a.first == b.first) {
    return a.second.first < b.second.first;
  } else
    return a.first < b.first;
}
bool isprime(long long x) {
  if (x == 1) return false;
  if (x == 2 || x == 3 || x == 5) return true;
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  for (long long t = 0; t < T; t++) {
    long long n, k;
    cin >> n >> k;
    multiset<long long> all;
    for (long long i = 0; i < n; i++) {
      long long a;
      cin >> a;
      all.insert(a);
    }
    auto a = all.begin();
    if (k > 1)
      while (a != all.end() and all.size() > 1) {
        long long x = (*a) * (k);
        all.erase(x);
        a++;
      }
    cout << all.size();
  }
  return 0;
}
