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;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  long long int n, m, i, j;
  cin >> n >> m;
  long long int arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  long long int mx = *max_element(arr, arr + n);
  vector<long long int> adj[mx + 1];
  for (i = 0; i < n; i++) {
    long long int x = arr[i];
    long long int i = 0;
    while (x != 0) {
      adj[x].push_back(i);
      x /= 2;
      i++;
    }
  }
  long long int ans = 100000000;
  for (i = 1; i <= mx; i++) {
    if (adj[i].size() != 0) {
      sort(adj[i].begin(), adj[i].end());
    }
  }
  for (i = 1; i <= mx; i++) {
    if (adj[i].size() >= m) {
      long long int x = 0;
      for (j = 0; j < m; j++) {
        x += adj[i][j];
      }
      ans = min(ans, x);
    }
  }
  cout << ans << '\n';
}