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;
const int maxn = 2e5 + 5;
long long a[maxn];
long long b[maxn];
map<long long, int> m;
int main() {
  int n, k;
  cin >> n >> k;
  int flag = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    m[a[i]]++;
    if (m[a[i]] >= k) flag = 1;
  }
  if (flag) {
    cout << 0 << endl;
    return 0;
  }
  sort(a + 1, a + 1 + n);
  long long minn = 1e16;
  for (int i = 1; i <= n; i++) {
    b[i] = b[i - 1] + a[i];
  }
  for (int i = 1; i <= n; i++) {
    minn =
        min(minn, (-b[i] + i * a[i]) - (n - k) + b[n] - b[i] - (n - i) * a[i]);
    if (i >= k) minn = min(minn, (-b[i] + i * a[i]) - (i - k));
    if (i + k <= n + 1)
      minn = min(minn, b[n] - b[i - 1] - (n - i + 1) * a[i] - (n - i + 1 - k));
  }
  cout << minn << endl;
  return 0;
}
