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);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<vector<int> > vals(200 * 1000 + 11);
  for (int i = 0; i < n; ++i) {
    int x = a[i];
    int cur = 0;
    while (x > 0) {
      vals[x].push_back(cur);
      x /= 2;
      ++cur;
    }
  }
  int ans = 1e9;
  for (int i = 0; i <= 200 * 1000; ++i) {
    sort(vals[i].begin(), vals[i].end());
    if (int(vals[i].size()) < k) continue;
    ans = min(ans, accumulate(vals[i].begin(), vals[i].begin() + k, 0));
  }
  cout << ans << endl;
  return 0;
}