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;
int n, k;
long long fw[200005], bk[200005];
int af[200005], ab[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> k;
  map<int, int> mp;
  for (int i = 0, x; i < n; i++) cin >> x, mp[x]++;
  vector<pair<int, int>> b(mp.begin(), mp.end());
  for (auto [x, y] : b)
    if (y >= k) return cout << "0\n", 0;
  int m = b.size();
  long long cost = 0;
  int amt = 0;
  for (int i = 0; i < m - 1; i++) {
    fw[i] = cost;
    amt += b[i].second;
    af[i] = amt;
    cost += 1ll * (b[i + 1].first - b[i].first) * amt;
  }
  cost = 0;
  amt = 0;
  for (int i = m - 1; i > 0; i--) {
    bk[i] = cost;
    amt += b[i].second;
    ab[i] = amt;
    cost += 1ll * (b[i].first - b[i - 1].first) * amt;
  }
  long long sol = 1e18;
  for (int i = 0; i < m; i++) {
    if (i > 0 && af[i - 1] + b[i].second >= k)
      sol =
          min(sol, fw[i - 1] + (b[i].first - b[i - 1].first - 1ll) * af[i - 1] +
                       k - b[i].second);
    if (i + 1 < m && ab[i + 1] + b[i].second >= k)
      sol =
          min(sol, bk[i + 1] + (b[i + 1].first - b[i].first - 1ll) * ab[i + 1] +
                       k - b[i].second);
    if (i > 0 && i + 1 < m) {
      sol = min(
          sol, fw[i - 1] + (b[i].first - b[i - 1].first - 1ll) * af[i - 1] +
                   bk[i + 1] + (b[i + 1].first - b[i].first - 1ll) * ab[i + 1] +
                   k - b[i].second);
    }
  }
  cout << sol << '\n';
}