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 = (int)2e5 + 9;
long long Cl[N];
long long Cr[N];
int sl[N];
int sr[N];
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  vector<pair<int, int> > cmp;
  for (int i = 0; i < n; i++) {
    if (cmp.empty() || a[i] != cmp.back().first) {
      cmp.push_back(make_pair(a[i], 1));
    } else {
      cmp.back().second++;
    }
  }
  int m = cmp.size();
  for (int i = 0; i < m; i++) {
    if (cmp[i].second >= k) {
      cout << "0\n";
      return 0;
    }
  }
  Cl[0] = 0;
  sl[0] = cmp[0].second;
  for (int i = 1; i < m; i++) {
    sl[i] = sl[i - 1] + cmp[i].second;
    Cl[i] = Cl[i - 1] + (sl[i - 1] * 1ll * (cmp[i].first - cmp[i - 1].first));
  }
  Cr[m - 1] = 0;
  sr[m - 1] = cmp[m - 1].second;
  for (int i = m - 2; i >= 0; i--) {
    sr[i] = sr[i + 1] + cmp[i].second;
    Cr[i] = Cr[i + 1] + (sr[i + 1] * 1ll * (cmp[i + 1].first - cmp[i].first));
  }
  long long ans = (long long)1e18;
  long long cost;
  for (int i = 1; i + 1 < m; i++) {
    cost = Cl[i - 1] + Cr[i + 1];
    cost += (cmp[i + 1].first - cmp[i].first - 1) * 1ll * sr[i + 1];
    cost += (cmp[i].first - cmp[i - 1].first - 1) * 1ll * sl[i - 1];
    cost += k - cmp[i].second;
    ans = min(ans, cost);
  }
  for (int i = 1; i < m; i++) {
    if (sl[i] >= k) {
      cost = Cl[i - 1];
      cost += (cmp[i].first - cmp[i - 1].first - 1) * 1ll * sl[i - 1];
      cost += k - cmp[i].second;
      ans = min(ans, cost);
    }
  }
  for (int i = m - 2; i >= 0; i--) {
    if (sr[i] >= k) {
      cost = Cr[i + 1];
      cost += (cmp[i + 1].first - cmp[i].first - 1) * 1ll * sr[i + 1];
      cost += k - cmp[i].second;
      ans = min(ans, cost);
      break;
    }
  }
  cout << ans << "\n";
  return 0;
}