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::sync_with_stdio(false), cin.tie(0);
  ;
  long long n, k, i;
  cin >> n >> k;
  long long a[n], pr[n + 1];
  map<long long, long long> f;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  int fl = 0;
  for (auto e : f) {
    if (e.second >= k) {
      cout << "0\n";
      fl = 1;
      break;
    }
  }
  if (!fl) {
    sort(a, a + n);
    pr[0] = a[0];
    for (i = 1; i < n; i++) pr[i] = pr[i - 1] + a[i];
    long long ans = 1e18 + 5;
    long long left, right, ex, costl, costr;
    for (i = 0; i < n; i++) {
      left = lower_bound(a, a + n, a[i]) - a;
      right = n - (upper_bound(a, a + n, a[i]) - a);
      ex = n - left - right;
      costl = left * (a[i] - 1);
      if (left) costl -= pr[left - 1];
      costr = pr[n - 1] - right * (a[i] + 1);
      if ((n - right - 1) >= 0) costr -= pr[n - right - 1];
      if (left + ex >= k) ans = min(ans, costl + k - ex);
      if (right + ex >= k) ans = min(ans, costr + k - ex);
      ans = min(ans, costl + costr + k - ex);
    }
    cout << ans << '\n';
  }
}