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;
using namespace std::chrono;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long l[n], r[n];
  l[0] = r[n - 1] = 0;
  long long bst = 1000000000000000000;
  for (long long i = 1; i < n; i++)
    l[i] = l[i - 1] + i - 1 + i * (a[i] - a[i - 1] - 1);
  for (long long i = n - 2; i >= 0; i--)
    r[i] = r[i + 1] + n - i - 2 + (n - i - 1) * (a[i + 1] - a[i] - 1);
  for (long long i = 0; i < n; i++) {
    long long j = i;
    while (j < n and a[j] == a[i]) j++;
    j--;
    long long len = j - i + 1;
    long long ans;
    if (len >= k)
      ans = 0;
    else {
      if (i >= k - len and n - j - 1 >= k - len)
        ans = min(l[i], r[j]);
      else if (i >= k - len)
        ans = l[i];
      else if (n - j - 1 >= k - len)
        ans = r[j];
      else
        ans = l[i] + r[j];
      ans += k - len;
    }
    bst = min(bst, ans);
    i = j;
  }
  cout << bst << endl;
}