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(0);
  cin.tie(0);
  const long long oo = 1e18;
  int n, k;
  cin >> n >> k;
  vector<long long> v(n + 1);
  for (int i = 1; i <= n; i++) cin >> v[i];
  sort(v.begin() + 1, v.end());
  vector<long long> acum(n + 1);
  for (int i = 1; i <= n; i++) acum[i] = acum[i - 1] + v[i];
  long long ans = oo;
  int last = 0;
  for (int i = 1; i <= n; i++) {
    if (v[i] != v[i + 1] || i == n) {
      int many = i - last;
      if (many >= k) {
        ans = 0;
        break;
      }
      long long left = (v[i] - 1ll) * last - acum[last];
      long long right = acum[n] - acum[i] - (v[i] + 1ll) * 1ll * (n - i);
      if (i >= k) ans = min(ans, left + 1ll * k - 1ll * many);
      if (n >= k + last) ans = min(ans, right + 1ll * k - 1ll * many);
      ans = min(ans, left + right + 1ll * k - 1ll * many);
      last = i;
    }
  }
  cout << ans << '\n';
  return 0;
}