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;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  vector<long long> pre(n);
  pre[0] = a[0];
  for (int i = 1; i < n; i++) pre[i] = a[i] + pre[i - 1];
  vector<int> pos;
  for (int i = 0; i < n - 1; i++)
    if (a[i] != a[i + 1]) pos.push_back(i);
  pos.push_back(n - 1);
  long long ans = 1e18;
  for (int i = 0; i < pos.size(); i++) {
    int lpos = (i ? pos[i - 1] + 1 : 0);
    int rpos = pos[i];
    long long need = k - (rpos - lpos + 1);
    if (need <= 0) {
      cout << 0;
      return 0;
    }
    long long lsum = (lpos * a[lpos]) - (lpos ? pre[lpos - 1] : 0);
    long long rsum = (pre[n - 1] - pre[rpos]) - (n - rpos - 1) * a[lpos];
    if (lpos >= need) ans = min(ans, lsum - lpos + need);
    if (n - rpos - 1 >= need) ans = min(ans, rsum - (n - rpos - 1) + need);
    ans = min(ans, lsum + rsum - (n - k));
  }
  cout << ans;
  return 0;
}