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;
int main() {
  long long n, k, i, j, ans;
  cin >> n >> k;
  vector<long long> v(n);
  for (auto &it : v) {
    cin >> it;
  }
  sort(v.begin(), v.end());
  long long pre[n + 1];
  pre[0] = 0;
  for (i = 0; i < n; i++) {
    pre[i + 1] = pre[i] + v[i];
  }
  ans = LLONG_MAX;
  for (i = 0; i < n; i = j) {
    j = upper_bound(v.begin(), v.end(), v[i]) - v.begin();
    if (j - i >= k) {
      ans = 0;
      break;
    }
    if (i >= k - j + i) {
      ans = min(ans, k - j + i + ((v[i] - 1) * i) - pre[i]);
    }
    if (n - j >= k - j + i) {
      ans = min(ans, k - j + i - ((v[i] + 1) * (n - j)) + pre[n] - pre[j]);
    }
    ans = min(ans, k - j + i + ((v[i] - 1) * i) - pre[i] -
                       ((v[i] + 1) * (n - j)) + pre[n] - pre[j]);
  }
  cout << ans;
  return 0;
}
