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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  while (TESTS--) {
    int i, n, j, k;
    cin >> n >> k;
    vector<int> v(n);
    vector<vector<int> > a(200005);
    for (i = 0; i < n; i++) cin >> v[i];
    int ans = INT_MAX;
    for (i = 0; i < n; i++) {
      int t = v[i], s = 0;
      while (t) {
        a[t].push_back(s);
        s++;
        t /= 2;
      }
    }
    for (i = 0; i < 200005; i++) {
      if (a[i].size() >= k) {
        sort((a[i]).begin(), (a[i]).end());
        int sum = 0;
        for (j = 0; j < k; j++) sum += a[i][j];
        ans = min(ans, sum);
      }
    }
    cout << ans << endl;
  }
}