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;
template <typename T>
T maxelement(T x, T y) {
  return (x > y) ? x : y;
}
long long a[500005];
map<long long, long long> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, j, k, l, n;
  cin >> n >> k;
  set<long long> s;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]]++;
    s.insert(a[i]);
  }
  vector<long long> elements(s.size() + 1, 0);
  long long st = 1;
  for (auto x : s) {
    elements[st++] = x;
  }
  vector<long long> sum(s.size() + 1, 0);
  vector<long long> fre(s.size() + 1, 0);
  long long sz = s.size();
  for (i = 1; i <= sz; i++) {
    fre[i] = fre[i - 1] + mp[elements[i]];
    sum[i] = sum[i - 1] + mp[elements[i]] * elements[i];
  }
  long long ans = 1e18;
  for (i = 1; i <= sz; i++) {
    long long req = max(0LL, k - mp[elements[i]]);
    if (req == 0) {
      cout << "0";
      return 0;
    }
    if (fre[i - 1] >= req || ((fre[sz] - fre[i]) >= req)) {
      long long Leftmini = 1e17;
      long long Rightmini = 1e17;
      if (fre[i - 1] >= req) {
        Leftmini = min(Leftmini, fre[i - 1] * (elements[i] - 1) - sum[i - 1]);
        Leftmini += req;
      }
      if (fre[sz] - fre[i] >= req) {
        Rightmini = min(Rightmini, (sum[sz] - sum[i] -
                                    (fre[sz] - fre[i]) * (elements[i] + 1)));
        Rightmini += req;
      }
      ans = min(ans, min(Leftmini, Rightmini));
    } else {
      long long Leftmini = 0;
      long long Rightmini = 0;
      Leftmini = fre[i - 1] * (elements[i] - 1) - sum[i - 1];
      Rightmini = sum[sz] - sum[i] - (fre[sz] - fre[i]) * (elements[i] + 1);
      ans = min(ans, Leftmini + Rightmini + req);
    }
  }
  cout << max(ans, 0LL) << "\n";
}