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;
vector<int> v[200005];
int arr[200005];
pair<int, int> fr[200005];
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    while (x) {
      v[i].push_back(x);
      x /= 2;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < int((v[i]).size()); j++) {
      if (fr[v[i][j]].first < k) {
        fr[v[i][j]].first++;
        fr[v[i][j]].second += j;
      }
    }
  }
  int ans = (1 << 30);
  for (int i = 0; i < 200005; i++) {
    if (fr[i].first == k) ans = min(ans, fr[i].second);
  }
  cout << ans << endl;
  return 0;
}