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;
const long long LL_INF = (long long)2e18 + 5;
const int SIZE = 1e6 + 10;
long long n, k, K, res = LL_INF;
long long a[SIZE];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> K;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int tar = 0; tar <= a[n - 1]; tar++) {
    for (int i = 0; i < n; i++) {
      if (a[i] < tar) continue;
      int j = i;
      long long op = 0;
      k = K;
      for (j; k && j < n; j++) {
        if (a[j] == tar)
          k--;
        else {
          long long cnt = 0;
          long long tmp = a[j];
          while (tmp > tar) {
            tmp /= 2;
            cnt++;
          }
          if (tmp == tar) {
            op += cnt;
            k--;
          }
        }
      }
      if (k == 0) res = min(res, op);
      break;
    }
  }
  cout << res;
  return 0;
}