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 main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  set<int> av;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int w = a[i];
    av.insert(w);
    while (w) {
      w >>= 1;
      av.insert(w);
    }
  }
  vector<int> u;
  for (auto q : av) u.push_back(q);
  vector<vector<int> > t(n, vector<int>(av.size()));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < av.size(); j++) {
      if (a[i] < u[j])
        t[i][j] = 1e9;
      else {
        t[i][j] = 1e9;
        int w = a[i];
        if (u[j] == w) {
          t[i][j] = 0;
        }
        int cur = 0;
        while (w) {
          w >>= 1;
          cur++;
          if (u[j] == w) t[i][j] = cur;
        }
      }
    }
  }
  t[0][0] = t[0][0];
  long long res = 1e9;
  for (int i = 0; i < av.size(); i++) {
    vector<int> nw;
    for (int j = 0; j < n; j++) nw.push_back(t[j][i]);
    sort(nw.begin(), nw.end());
    bool f = false;
    for (int j = 0; j < k; j++)
      if (nw[j] == 1e9) {
        f = true;
      }
    if (!f) {
      long long q = 0;
      for (int j = 0; j < k; j++) q += nw[j];
      res = min(res, q);
    }
  }
  cout << res;
  cin >> res;
}