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;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int> > arr;
  arr.reserve(20);
  vector<int> temp(n);
  for (int i = 0; i < n; i++) cin >> temp[i];
  sort(temp.begin(), temp.end());
  arr.push_back(temp);
  bool is = true;
  int i = 0;
  while (is) {
    is = false;
    for (int j = 0; j < n; j++) {
      temp[j] = arr[i][j] / 2;
      if (temp[j]) is = true;
    }
    arr.push_back(temp);
    i++;
  }
  int ans = 0;
  int temp_ans;
  int fre, p, l = arr.size(), fre_temp;
  int mx = *max_element(arr[0].begin(), arr[0].end());
  for (int i = 0; i < k; i++) ans += log2(arr[0][i]) + 1;
  for (int i = 1; i <= mx; i++) {
    temp_ans = 0;
    fre = 0, fre_temp = 0;
    p = 0;
    for (int j = 0; j < l; j++) {
      fre_temp = upper_bound(arr[j].begin(), arr[j].end(), i) -
                 lower_bound(arr[j].begin(), arr[j].end(), i);
      if (fre + fre_temp >= k) {
        temp_ans += p * (k - fre);
        fre = k;
        break;
      }
      fre += fre_temp;
      temp_ans += p * fre_temp;
      p++;
    }
    if (fre >= k) ans = min(ans, temp_ans);
  }
  cout << ans << "\n";
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}