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);
  long long n, m;
  cin >> n >> m;
  vector<long long> theval(n, 0);
  long long sum = 0, output = 100000000000000000, leen = 1;
  vector<long long> prefix(n + 10, 0);
  for (int i = 0; i < n; i++) {
    cin >> theval[i];
    sum += theval[i];
  }
  sort(theval.begin(), theval.end());
  for (int i = 1; i < n; i++) {
    if (theval[i] == theval[i - 1])
      leen++;
    else
      leen = 1;
    if (leen >= m) {
      cout << 0 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    prefix[i] += theval[i];
    if (i == 0) continue;
    prefix[i] += prefix[i - 1];
  }
  for (int i = 0; i < n; i++) {
    long long ans1 = 100000000000000000, ans2 = 100000000000000000,
              ans3 = 100000000000000000;
    if (i >= m - 1) {
      ans1 = (i * (theval[i] - 1) - (prefix[i] - theval[i])) + m - 1;
    }
    if (n - i - 1 >= m - 1) {
      ans2 = (sum - prefix[i] - (n - i - 1) * (theval[i] + 1)) + m - 1;
    }
    ans3 = (i * (theval[i] - 1) - (prefix[i] - theval[i])) +
           (sum - prefix[i] - (n - i - 1) * (theval[i] + 1)) + m - 1;
    output = min(output, min(ans1, min(ans2, ans3)));
  }
  cout << output << endl;
}