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(0);
  int n;
  int k;
  cin >> n >> k;
  int ar[n];
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  sort(ar, ar + n);
  long long lista[n];
  int cantidad[n];
  int index = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      lista[i] = ar[i];
      cantidad[index] = 1;
    } else if (ar[i] == ar[i - 1]) {
      cantidad[index]++;
    } else {
      index++;
      cantidad[index] = 1;
      lista[index] = ar[i];
    }
  }
  int diferentes = index + 1;
  long long L[diferentes];
  long long R[diferentes];
  int LeftAmount[diferentes];
  int RightAmount[diferentes];
  L[0] = 0;
  LeftAmount[0] = 0;
  R[diferentes - 1] = 0;
  RightAmount[diferentes - 1] = 0;
  for (int i = 1; i < diferentes; i++) {
    LeftAmount[i] = LeftAmount[i - 1] + cantidad[i - 1];
    L[i] = L[i - 1] + (lista[i] - 1 - lista[i - 1]) * cantidad[i - 1];
    L[i] += (lista[i] - 1 - (lista[i - 1] - 1)) * LeftAmount[i - 1];
  }
  for (int i = diferentes - 2; i >= 0; i--) {
    RightAmount[i] = RightAmount[i + 1] + cantidad[i + 1];
    R[i] = R[i + 1] + (lista[i + 1] - (lista[i] + 1)) * cantidad[i + 1];
    R[i] += (lista[i + 1] - lista[i]) * RightAmount[i + 1];
  }
  long long ans = 1000000000000000005;
  for (int i = 0; i < diferentes; i++) {
    if (cantidad[i] >= k) {
      ans = 0;
    } else {
      int needed = k - cantidad[i];
      if ((LeftAmount[i] >= needed) && (RightAmount[i] >= needed)) {
        ans = min(min(L[i], R[i]) + needed, ans);
      } else if (LeftAmount[i] >= needed) {
        ans = min(L[i] + needed, ans);
      } else if (RightAmount[i] >= needed) {
        ans = min(R[i] + needed, ans);
      } else {
        ans = min(L[i] + R[i] + needed, ans);
      }
    }
  }
  cout << ans;
  return 0;
}