#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> freq_op(200005, vector<int>(2, 0));
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    freq_op[x][0]++;
    int op = 0;
    while (x > 0) {
      x /= 2;
      op++;
      if (freq_op[x][0] < k) {
        freq_op[x][0]++;
        freq_op[x][1] += op;
      }
    }
  }
  int minimum = INT_MAX;
  for (int i = 0; i < freq_op.size(); i++) {
    if (freq_op[i][0] >= k) minimum = min(minimum, freq_op[i][1]);
  }
  cout << minimum;
  return 0;
}

### Prompt

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

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 main() {
  int n, k;
  cin >> n >> k;
  vector<vector<int>> freq_op(200005, vector<int>(2, 0));
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    freq_op[x][0]++;
    int op = 0;
    while (x > 0) {
      x /= 2;
      op++;
      if (freq_op[x][0] < k) {
        freq_op[x][0]++;
        freq_op[x][1] += op;
      }
    }
  }
  int minimum = INT_MAX;
  for (int i = 0; i < freq_op.size(); i++) {
    if (freq_op[i][0] >= k) minimum = min(minimum, freq_op[i][1]);
  }
  cout << minimum;
  return 0;
}

```