#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-11;
const double pi = acos(-1.0);
const int p = 1000000007;
int countbit(int n) { return n == 0 ? 0 : countbit(n / 2) + n % 2; }
int lowbit(int n) { return (n ^ (n - 1)) & n; }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long toInt64(string s) {
  istringstream sin(s);
  long long t;
  sin >> t;
  return t;
}
string toStr(long long t) {
  stringstream sin;
  sin << t;
  string s;
  sin >> s;
  return s;
}
template <typename T>
T sqr(T n) {
  return n * n;
}
template <typename T>
void out(T& v) {
  for (int i = 0; i < v.size(); i++)
    cout << v[i] << (i == v.size() - 1 ? '\n' : ' ');
}
class Timer {
 public:
  double start;
  Timer() { start = static_cast<double>(clock()); }
  void out() { cout << "TIME: " << (clock() - start) / CLOCKS_PER_SEC << endl; }
};
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n), ok(n, true);
  map<long long, long long> index;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort((a).begin(), (a).end());
  for (long long i = 0; i < n; i++) {
    index.insert(pair<long long, long long>(a[i], i));
  }
  long long result = 0;
  for (long long i = 0; i < n; ++i)
    if (ok[i]) {
      result++;
      if (index.find(a[i] * k) != index.end()) {
        ok[index[a[i] * k]] = false;
      }
    }
  cout << result << endl;
}
int main() {
  cout.sync_with_stdio(false);
  cin.sync_with_stdio(false);
  solve();
  return EXIT_SUCCESS;
}

### Prompt

Please create a solution in CPP to the following problem:

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}.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-11;
const double pi = acos(-1.0);
const int p = 1000000007;
int countbit(int n) { return n == 0 ? 0 : countbit(n / 2) + n % 2; }
int lowbit(int n) { return (n ^ (n - 1)) & n; }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long toInt64(string s) {
  istringstream sin(s);
  long long t;
  sin >> t;
  return t;
}
string toStr(long long t) {
  stringstream sin;
  sin << t;
  string s;
  sin >> s;
  return s;
}
template <typename T>
T sqr(T n) {
  return n * n;
}
template <typename T>
void out(T& v) {
  for (int i = 0; i < v.size(); i++)
    cout << v[i] << (i == v.size() - 1 ? '\n' : ' ');
}
class Timer {
 public:
  double start;
  Timer() { start = static_cast<double>(clock()); }
  void out() { cout << "TIME: " << (clock() - start) / CLOCKS_PER_SEC << endl; }
};
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n), ok(n, true);
  map<long long, long long> index;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort((a).begin(), (a).end());
  for (long long i = 0; i < n; i++) {
    index.insert(pair<long long, long long>(a[i], i));
  }
  long long result = 0;
  for (long long i = 0; i < n; ++i)
    if (ok[i]) {
      result++;
      if (index.find(a[i] * k) != index.end()) {
        ok[index[a[i] * k]] = false;
      }
    }
  cout << result << endl;
}
int main() {
  cout.sync_with_stdio(false);
  cin.sync_with_stdio(false);
  solve();
  return EXIT_SUCCESS;
}

```