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() {
  int n, k, x, co;
  cin >> n >> k;
  vector<pair<int, int>> li(200001, make_pair(0, 0));
  vector<int> da(n);
  for (int i = (0); i < (int)(n); i++) {
    cin >> da[i];
    li[da[i]].first++;
    if (li[da[i]].first > k - 1) {
      cout << 0 << endl;
      return 0;
    }
  }
  bool ju = true;
  co = 0;
  int ans = INT_MAX;
  while (ju) {
    co++;
    ju = false;
    for (int i = (0); i < (int)(n); i++) {
      if (da[i]) {
        da[i] /= 2;
        li[da[i]].first++;
        li[da[i]].second += co;
        ju = true;
        if (li[da[i]].first > k - 1) {
          ans = min(ans, li[da[i]].second);
        }
      }
    }
  }
  cout << ans << endl;
}