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;
string sfrom(string s, int i, int j) { return s.substr(i, j - i + 1); }
template <typename reservedArgs>
string to_string(reservedArgs args) {
  stringstream sso;
  sso << args;
  return sso.str();
}
template <typename reservedArgs>
string to_string(reservedArgs args, int precision) {
  stringstream sso;
  sso << fixed << setprecision(precision);
  sso << args;
  return sso.str();
}
long long hell = 1000000007LL;
template <typename reservedArgs>
reservedArgs gcd(reservedArgs n1, reservedArgs n2) {
  return (n2 == 0 ? n1 : gcd(n2, n1 % n2));
}
class compare {
 public:
  bool operator()(int lowPriority, int highPriority) {
    return lowPriority < highPriority;
  }
};
priority_queue<int, vector<int>, compare> pq;
const int N = 2e5 + 5;
vector<int> dist[N];
void main1() {
  int n, k, a;
  cin >> n >> k;
  for (int(i1) = 0; (i1) < (n); (i1)++) {
    cin >> a;
    for (int i = 0; true; i++, a /= 2) {
      dist[a].push_back(i);
      if (a == 0) break;
    }
  }
  int ans = INT_MAX, tmp;
  for (int(i) = 0; (i) < (N); (i)++) {
    if (dist[i].size() >= k) {
      sort(dist[i].begin(), dist[i].end());
      tmp = 0;
      for (int(j) = 0; (j) < (k); (j)++) tmp += dist[i][j];
      ans = min(ans, tmp);
    }
  }
  cout << ans << endl;
}
int main() {
  int t = 1;
  for (int(i1) = 0; (i1) < (t); (i1)++) {
    main1();
  }
  return 0;
}