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>
#pragma GCC optimize("03")
using namespace std;
long long int const mod = 1000000007;
long long int br[200100][61];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k, x, ans = 1e12, mx = 0, cnt, k1, d;
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) {
    cin >> x;
    mx = max(mx, x);
    cnt = 0;
    while (x >= 1) {
      br[x][cnt] += 1;
      x /= 2;
      cnt += 1;
    }
  }
  for (long long int i = 0; i <= mx; i++) {
    k1 = k;
    cnt = 0;
    for (long long int j = 0; j <= 60 && k1 > 0; j++) {
      d = min(k1, br[i][j]);
      cnt += d * j;
      k1 -= d;
    }
    if (k1 == 0) ans = min(ans, cnt);
  }
  cout << ans << "\n";
  return 0;
}