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;
void solve() {
  int n, k;
  cin >> n >> k;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  int ops[200001] = {0};
  int count[200001] = {0};
  for (int i = 0; i < n; i++) {
    int x = arr[i];
    int op_count = 0;
    while (x >= 1) {
      if (count[x] < k) {
        ops[x] += op_count;
        count[x]++;
      }
      x = (x / 2);
      op_count++;
    }
  }
  int ans = ops[1];
  for (int i = 1; i <= 200000; i++) {
    if (count[i] >= k) ans = min(ans, ops[i]);
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << std::fixed;
  cout << std::setprecision(10);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}