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, p[200005], s[200005], a[200005];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = 0; i <= n - k; ++i) {
    if (a[i] == a[i + k - 1]) return cout << 0, 0;
  }
  p[0] = a[0];
  for (int i = 1; i < n; ++i) {
    p[i] = a[i] + p[i - 1];
  }
  s[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; --i) {
    s[i] = a[i] + s[i + 1];
  }
  long long j = 0, ans = LLONG_MAX;
  for (int i = 0; i < n;) {
    while (j < n && a[j] == a[i]) j++;
    long long pans = 0, sans = 0, cnt = (j - i);
    if (i == 0)
      pans = LLONG_MAX;
    else {
      pans = i * (a[i] - 1) - p[i - 1];
      pans += min(k - cnt, (long long)i);
      if (j < k) {
        pans += s[j] - (n - j) * (a[i] + 1);
        pans += (k - j);
      }
    }
    if (j == n)
      sans = LLONG_MAX;
    else {
      sans = s[j] - (n - j) * (a[i] + 1);
      sans += min(n - j, k - (j - i));
      if (n - i < k) {
        sans += i * (a[i] - 1) - p[i - 1];
        sans += (k - (n - i));
      }
    }
    ans = min(ans, min(sans, pans));
    i = j;
  }
  cout << ans;
  return 0;
}