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
#include <bits/stdc++.h>
using namespace std;
const long long N = 200010;
long long a[N];
long long sum[N];
long long c[N];
long long t;
signed main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  long long res = 1e18;
  for (long long i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + a[i];
  }
  for (long long i = 1; i <= n; i++) {
    long long j = i;
    while (j <= n && a[j] == a[i]) j++;
    long long len = j - i;
    if (len >= k) res = 0;
    long long l = -sum[i - 1] + (i - 1) * a[i];
    long long r = sum[n] - sum[j - 1] - (n - j + 1) * a[i];
    if (j - 1 >= k)
      res = min(res, l - (j - 1 - k));
    else
      res = min(res, l + r - (n - k));
    if (n - i + 1 >= k) res = min(res, r - (n - i + 1 - k));
    i = j - 1;
  }
  res = max(res, 0ll);
  cout << res << endl;
  return 0;
}
