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;
ll n, k;
ll v[200002];
ll s[200001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> v[i];
  sort(v + 1, v + 1 + n);
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + v[i];
  ll ans = 1e14;
  for (ll i = 1; i <= n; i++) {
    ll j = i;
    while (v[j] == v[j + 1]) j++;
    ll fr = j - i + 1;
    if (fr >= k) {
      cout << "0";
      return 0;
    }
    if (i - 1 >= k - fr) {
      ll cost = (i - 1) * (v[i] - 1) - s[i - 1];
      cost += k - fr;
      ans = min(ans, cost);
    }
    if (n - j >= k - fr) {
      ll cost = s[n] - s[j] - (n - j) * (v[i] + 1);
      cost += k - fr;
      ans = min(ans, cost);
    }
    ll cost = (s[n] - s[j] - (n - j) * (v[i] + 1)) +
              ((i - 1) * (v[i] - 1) - s[i - 1]);
    cost += k - fr;
    ans = min(ans, cost);
    i = j;
  }
  cout << ans;
  return 0;
}