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;
using ll = long long;
int main() {
  cin.sync_with_stdio(false), cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a.begin() + 1, a.end());
  vector<ll> sum(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i];
  }
  ll res = sum[n];
  for (int l = 1, r; l <= n; l = r + 1) {
    r = l;
    while (r + 1 <= n && a[r] == a[r + 1]) {
      ++r;
    }
    if (r - l + 1 >= k) {
      res = 0;
      break;
    }
    int need_l = min(l - 1, k - (r - l + 1)), need_r = k - (r - l + 1) - need_l;
    if (need_r <= n - r) {
      ll current = 0;
      if (need_l) {
        current += (ll)(l - 1) * (a[l] - 1) - sum[l - 1] + need_l;
      }
      if (need_r) {
        current += (sum[n] - sum[r]) - (ll)(n - r) * (a[r] + 1) + need_r;
      }
      res = min(res, current);
    }
    need_r = min(n - r, k - (r - l + 1)), need_l = k - (r - l + 1) - need_r;
    if (need_l <= l - 1) {
      ll current = 0;
      if (need_l) {
        current += (ll)(l - 1) * (a[l] - 1) - sum[l - 1] + need_l;
      }
      if (need_r) {
        current += (sum[n] - sum[r]) - (ll)(n - r) * (a[r] + 1) + need_r;
      }
      res = min(res, current);
    }
  }
  cout << res << "\n";
}