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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
const ll INF = 1e18;
int n, k;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> k;
  vi v(n);
  for (int& x : v) cin >> x;
  sort(begin(v), end(v));
  vector<ll> a(n), b(n);
  for (int i = 0; i < n; ++i) a[i] = b[i] = v[i];
  for (int i = 1; i < n; ++i) a[i] += a[i - 1];
  for (int i = n - 2; i >= 0; --i) b[i] += b[i + 1];
  ll r = INF;
  for (int i = 0; i < n;) {
    int j = i;
    while (j < n && v[i] == v[j]) ++j;
    if (j - i >= k) {
      r = 0;
      break;
    }
    int need = k - (j - i);
    int before = i;
    int after = n - j;
    ll icost = 1ll * before * v[i] - (i > 0 ? a[i - 1] : 0ll);
    ll jcost = (j < n ? b[j] : 0ll) - 1ll * after * v[i];
    if (need <= before) r = min(r, icost - (before - need));
    if (need <= after) r = min(r, jcost - (after - need));
    r = min(r, icost + jcost - (before + after - need));
    i = j;
  }
  cout << r << '\n';
}
