#include <bits/stdc++.h>
using namespace std;
int n, k;
int dem[200005];
int a[200006];
int kq[200006];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int buoc;
  int rs = 1000000000;
  for (int i = 0; i < n; i++) {
    buoc = 0;
    while (a[i] > 0) {
      dem[a[i]] += 1;
      kq[a[i]] += buoc;
      if (dem[a[i]] >= k) {
        rs = min(rs, kq[a[i]]);
      }
      buoc += 1;
      a[i] /= 2;
    }
  }
  cout << rs;
}

### Prompt

Create a solution in CPP for the following problem:

The only difference between easy and hard versions is the number of elements in the array.

You are given an array a consisting of n integers. In one move you can choose any a_i and divide it by 2 rounding down (in other words, in one move you can set a_i := ⌊(a_i)/(2)⌋).

You can perform such an operation any (possibly, zero) number of times with any a_i.

Your task is to calculate the minimum possible number of operations required to obtain at least k equal numbers in the array.

Don't forget that it is possible to have a_i = 0 after some operations, thus the answer always exists.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the i-th element of a.

Output

Print one integer — the minimum possible number of operations required to obtain at least k equal numbers in the array.

Examples

Input


5 3
1 2 2 4 5


Output


1


Input


5 3
1 2 3 4 5


Output


2


Input


5 3
1 2 3 3 3


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, k;
int dem[200005];
int a[200006];
int kq[200006];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int buoc;
  int rs = 1000000000;
  for (int i = 0; i < n; i++) {
    buoc = 0;
    while (a[i] > 0) {
      dem[a[i]] += 1;
      kq[a[i]] += buoc;
      if (dem[a[i]] >= k) {
        rs = min(rs, kq[a[i]]);
      }
      buoc += 1;
      a[i] /= 2;
    }
  }
  cout << rs;
}

```