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
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int n, k, arr[200005];
vector<int> states;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<vector<int> > sat(200005 + 30);
  for (int i = 0; i < n; i++) {
    int w = arr[i];
    int dev = 0;
    while (w > 0) {
      sat[w].push_back(dev);
      states.push_back(w);
      w /= 2;
      dev++;
    }
  }
  int A = INT_MAX;
  for (vector<int> kaz : sat) {
    if (kaz.size() >= k) {
      sort(kaz.begin(), kaz.end());
      int jemerends = 0;
      for (int i = 0; i < k; i++) {
        jemerends += kaz[i];
      }
      A = min(A, jemerends);
    }
  }
  cout << A << '\n';
}
