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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    long long int n, k, i, x;
    cin >> n >> k;
    long long int a[n];
    long long int aa[200005] = {0};
    priority_queue<long long int> Q[200005];
    for (i = 0; i < n; i++) {
      cin >> x;
      long long int cost = 0;
      while (x) {
        Q[x].push(cost);
        aa[x] += cost;
        if (Q[x].size() > k) {
          aa[x] -= Q[x].top();
          Q[x].pop();
        }
        cost++;
        x /= 2;
      }
      Q[x].push(cost);
      aa[x] += cost;
      if (Q[x].size() > k) {
        aa[x] -= Q[x].top();
        Q[x].pop();
      }
    }
    long long int ans = 1e18;
    for (i = 0; i < 200004; i++) {
      if (Q[i].size() >= k) ans = min(ans, aa[i]);
    }
    cout << ans << "\n";
  }
  return 0;
}
