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> am[2 * (int)1e5 + 1];
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  int in;
  for (int i = 0; i < n; i++) {
    cin >> in;
    int moves = 0;
    while (in != 0) {
      am[in].push_back(moves);
      in /= 2;
      moves += 1;
    }
    am[in].push_back(moves);
  }
  int ans = INT_MAX;
  for (int i = 0; i <= 2 * (int)1e5; i++) {
    if (am[i].size() >= k) {
      sort(am[i].begin(), am[i].end());
      int fin = 0;
      for (int j = 0; j < k; j++) fin += am[i][j];
      ans = min(ans, fin);
    }
  }
  printf("%d\n", ans);
  return 0;
}