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;
const int fx[] = {+1, -1, +0, +0};
const int fy[] = {+0, +0, +1, -1};
vector<long long int> v, v1, res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, c, d, e, f, i, j, x, k, cnt1 = 0, id, chk = 0, sum = 0,
                                              cnt2 = 0;
  string s1, s2;
  cin >> a >> b;
  for (i = 0; i < a; i++) {
    cin >> c;
    v.push_back(c);
  }
  sort(v.begin(), v.end());
  long long int mx = *max_element(v.begin(), v.end());
  for (i = 0; i <= mx; i++) {
    res.clear();
    for (j = 0; j < a; j++) {
      long long int n = v[j];
      cnt1 = 0;
      while (n > i) {
        n /= 2;
        cnt1++;
      }
      if (n == i) {
        res.push_back(cnt1);
      }
    }
    sort(res.begin(), res.end());
    if (res.size() >= b) {
      sum = 0;
      for (j = 0; j < b; j++) {
        sum += res[j];
      }
      v1.push_back(sum);
    }
  }
  cout << *min_element(v1.begin(), v1.end());
  return 0;
}
