#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1;
  while (b) {
    if (b % 2 == 1) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2;
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = mul(fact[i - 1], i, M);
  }
  return mul(mul(fact[n], modInverse(fact[r], M), M),
             modInverse(fact[n - r], M), M);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (auto& w : v) cin >> w;
  sort(v.begin(), v.end());
  map<long long, long long> mm;
  set<long long> ss;
  map<long long, long long> dir;
  for (long long i = 0; i < n; i++) {
    if (v[i] % k == 0) {
      if (ss.find(v[i] / k) != ss.end()) {
        mm[dir[v[i] / k]]++;
        dir[v[i]] = dir[v[i] / k];
        ss.erase(v[i] / k);
        ss.insert(v[i]);
      } else {
        mm[v[i]]++;
        dir[v[i]] = v[i];
        ss.insert(v[i]);
      }
    } else {
      ss.insert(v[i]);
      dir[v[i]] = v[i];
      mm[v[i]]++;
    }
  }
  long long ans = 0;
  for (auto it = mm.begin(); it != mm.end(); it++) {
    if (it->second > 0) {
      ans += (it->second + 1) / 2;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1;
  while (b) {
    if (b % 2 == 1) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2;
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = mul(fact[i - 1], i, M);
  }
  return mul(mul(fact[n], modInverse(fact[r], M), M),
             modInverse(fact[n - r], M), M);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (auto& w : v) cin >> w;
  sort(v.begin(), v.end());
  map<long long, long long> mm;
  set<long long> ss;
  map<long long, long long> dir;
  for (long long i = 0; i < n; i++) {
    if (v[i] % k == 0) {
      if (ss.find(v[i] / k) != ss.end()) {
        mm[dir[v[i] / k]]++;
        dir[v[i]] = dir[v[i] / k];
        ss.erase(v[i] / k);
        ss.insert(v[i]);
      } else {
        mm[v[i]]++;
        dir[v[i]] = v[i];
        ss.insert(v[i]);
      }
    } else {
      ss.insert(v[i]);
      dir[v[i]] = v[i];
      mm[v[i]]++;
    }
  }
  long long ans = 0;
  for (auto it = mm.begin(); it != mm.end(); it++) {
    if (it->second > 0) {
      ans += (it->second + 1) / 2;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

```