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;
long long n, k;
long long a[200010];
long long pre[200010];
long long suf[200010];
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  for (long long i = n; i >= 1; i--) suf[i] = suf[i + 1] + a[i];
  long long ans = a[k] * k - pre[k];
  ans = min(ans, suf[n - k + 1] - a[n - k + 1] * k);
  for (long long i = 1; i <= n; i++)
    if (a[i + 1] != a[i]) {
      long long x = i;
      while (a[x] == a[i]) x--;
      long long l = (a[i] - 1) * x - pre[x];
      long long r = suf[i + 1] - (n - i) * (a[i] + 1);
      long long need = k - (i - x);
      ans = min(ans, l + r + need);
      if (k <= i) ans = min(ans, l + need);
      if (need <= n - i) ans = min(ans, r + need);
      if (need <= 0) {
        printf("0");
        return 0;
      }
    }
  cout << ans;
  return 0;
}