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>
using namespace std;
int arr[400000];
int initialD[400000];
vector<vector<int> > vecc;
vector<int> toint[400000];
int main() {
  int n, k;
  cin >> n >> k;
  vecc.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    initialD[arr[i]]++;
    int tem = arr[i];
    while (tem) {
      vecc[i].push_back(tem);
      tem /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    int sez = vecc[i].size();
    for (int j = 0; j < sez; j++) {
      toint[vecc[i][j]].push_back(j);
    }
  }
  long long ans = 1000000000;
  for (int i = 0; i < 200001; i++) {
    sort(toint[i].begin(), toint[i].end());
    int sez = toint[i].size();
    if (sez >= k) {
      long long temp = 0;
      for (int j = 0; j < k; j++) {
        temp += toint[i][j];
      }
      ans = min(ans, temp);
    }
  }
  cout << ans << '\n';
}
