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
#include <bits/stdc++.h>
using namespace std;
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
bool isprime(int k) {
  bool is = 1;
  for (int i = 2; i <= sqrt(k); i++)
    if (k % i == 0) is = 0;
  return is;
}
vector<int> vc[200000 + 5];
int32_t main() {
  ;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    int t = 0, x = a[i];
    while (x > 0) {
      vc[x].emplace_back(t);
      x /= 2;
      t++;
    }
  }
  int ans = 2147483647;
  for (int i = 0; i < 200005; i++) {
    if (vc[i].size() < k) continue;
    sort(vc[i].begin(), vc[i].end());
    ans = min(ans, accumulate(vc[i].begin(), vc[i].begin() + k, 0));
  }
  cout << ans << "\n";
  return 0;
}
