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
#include <bits/stdc++.h>
using namespace std;
long long i, n, a[200005], b[200005], c[200005], d[200005], s, q[200005], ans,
    k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> q[i];
  sort(q + 1, q + n + 1);
  for (i = 1; i <= n; i++)
    if (q[i] > q[i - 1]) {
      s++;
      a[s] = q[i];
      b[s] = 1;
    } else
      b[s]++;
  n = s;
  for (i = 1; i <= n; i++) {
    c[i] = c[i - 1] + a[i] * b[i];
    d[i] = d[i - 1] + b[i];
  }
  ans = 1e16;
  for (i = 1; i <= n; i++) {
    if (b[i] >= k) {
      cout << 0;
      return 0;
    }
    if (d[i - 1] + b[i] >= k)
      ans = min(ans, d[i - 1] * (a[i] - 1) - c[i - 1] + k - b[i]);
    if (d[n] - d[i] + b[i] >= k)
      ans = min(ans, c[n] - c[i] - (d[n] - d[i]) * (a[i] + 1) + k - b[i]);
    if (d[i - 1] + b[i] + d[n] - d[i] >= k)
      ans = min(ans, d[i - 1] * (a[i] - 1) - c[i - 1] + c[n] - c[i] -
                         (d[n] - d[i]) * (a[i] + 1) + k - b[i]);
  }
  cout << ans;
}
