Problem: You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

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

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, h;
  cin >> n >> k;
  int ar[n];
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  sort(ar, ar + n);
  int siz = (int)log2(ar[n - 1]) + 10;
  unordered_map<int, int> freq[siz];
  int grid[siz][n];
  memset(grid, 0, sizeof(grid));
  for (int i = 0; i < n; i++) {
    grid[0][i] = ar[i];
    freq[0][grid[0][i]]++;
  }
  for (int i = 1; i < siz; i++) {
    for (int j = 0; j < n; j++) {
      grid[i][j] = grid[i - 1][j] / 2;
      freq[i][grid[i][j]]++;
    }
  }
  long long int min_op = 9999999999;
  for (int i = 1; i <= ar[n - 1]; i++) {
    long long int op = 0;
    long long int count = 0;
    for (int j = 0; j < siz; j++) {
      if (freq[j][i] >= k - count) {
        op += (k - count) * j;
        count = k;
      } else {
        op += freq[j][i] * j;
        count += freq[j][i];
      }
      if (count >= k) {
        break;
      }
    }
    if (count >= k) {
      if (op < min_op) min_op = op;
    }
  }
  cout << min_op << "\n";
}