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;
const long long int MOD = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  vector<long long int> all;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    long long int x = a[i];
    while (x > 0) {
      all.push_back(x);
      x = x / 2;
    }
  }
  sort(a, a + n);
  long long int ans = 1e9;
  for (auto x : all) {
    long long int c = 0;
    long long int count = 0;
    for (long long int i = 0; i < n; i++) {
      long long int y = a[i];
      if (x == y) {
        count++;
      } else {
        long long int t = 0;
        while (y > 0) {
          if (x == y) {
            count++;
            c += t;
            break;
          }
          t++;
          y = y / 2;
        }
      }
      if (count >= k) {
        ans = min(ans, c);
        break;
      }
    }
  }
  cout << ans << "\n";
  return 0;
}