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;
const int N = 5e5 + 11;
long long a[N], s[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + a[i];
  int pos = 1;
  long long ans = 2e18;
  for (int i = 1; i <= n; i++)
    if (a[i] != a[i + 1]) {
      int kol = i - pos + 1;
      if (kol >= k) {
        ans = 0;
        break;
      }
      if (i >= k) {
        long long res = 0;
        res += (a[i] - 1) * 1ll * (pos - 1) - s[pos - 1];
        res += k - kol;
        ans = min(ans, res);
      }
      if (kol + n - i >= k) {
        long long res = 0;
        res += s[n] - s[i] - (a[i] + 1) * 1ll * (n - i);
        res += k - kol;
        ans = min(ans, res);
      }
      long long res = 0;
      res += (a[i] - 1) * 1ll * (pos - 1) - s[pos - 1];
      res += s[n] - s[i] - (a[i] + 1) * 1ll * (n - i);
      res += k - kol;
      ans = min(ans, res);
      pos = i + 1;
    }
  cout << ans;
}