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 v[200002];
pair<int, int> vec[200002];
long long st[200002], dr[200002];
int cntst[200002], cntdr[200002];
int cnt1;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, cnt = 1, i;
  long long min1 = (long long)2e17;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> v[i];
  sort(v + 1, v + n + 1);
  for (i = 1; i <= n; i++) {
    if (v[i] != v[i + 1]) {
      vec[++cnt1] = {v[i], cnt};
      cnt = 0;
    }
    cnt++;
  }
  n = cnt1;
  for (i = 1; i <= n; i++)
    if (vec[i].second >= k) {
      cout << "0";
      return 0;
    }
  for (i = 1; i <= n; i++) {
    st[i] = st[i - 1] + 1LL * cntst[i - 1] * (vec[i].first - vec[i - 1].first);
    cntst[i] = cntst[i - 1] + vec[i].second;
    if (cntst[i] >= k) min1 = min(min1, st[i] - (cntst[i] - k));
  }
  for (i = n; i >= 1; i--) {
    dr[i] = dr[i + 1] + 1LL * cntdr[i + 1] * (vec[i + 1].first - vec[i].first);
    cntdr[i] = cntdr[i + 1] + vec[i].second;
    if (cntdr[i] >= k) min1 = min(min1, dr[i] - (cntdr[i] - k));
  }
  for (i = 1; i <= n; i++) {
    if (k > cntst[i])
      min1 = min(
          min1, st[i] + dr[i + 1] + (k - cntst[i]) +
                    1LL * cntdr[i + 1] * (vec[i + 1].first - vec[i].first - 1));
    else
      break;
  }
  for (i = n; i >= 1; i--) {
    if (k > cntdr[i])
      min1 = min(
          min1, st[i - 1] + dr[i] + (k - cntdr[i]) +
                    1LL * cntst[i - 1] * (vec[i].first - vec[i - 1].first - 1));
    else
      break;
  }
  cout << min1;
  return 0;
}