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 n, k, l;
int t[200002];
long long e, v, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> t[i];
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= n; i++) {
    if (t[i] != t[i - 1]) l = i - 1;
    if (i - l >= k) {
      cout << 0 << "\n";
      return 0;
    }
  }
  for (int i = 1; i < k; i++) e += t[k] - t[i];
  for (int i = k + 1; i <= n; i++)
    if (t[i] == t[k]) e--;
  for (int i = n; i > n - k + 1; i--) v += t[i] - t[n - k + 1];
  for (int i = n - k; i >= 1; i--)
    if (t[i] == t[n - k + 1]) v--;
  for (int i = 1; i <= n; i++) m += abs(t[i] - t[(n + 1) / 2]);
  m -= n - k;
  cout << min({e, v, m}) << "\n";
  return 0;
}