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() {
  long long int n, k, i, fans, ffans, ans, dd, zz, ans2;
  cin >> n >> k;
  long long int a[n + 5], s[n + 5], d[n + 5];
  map<long long int, long long int> m;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    if (i == 0)
      s[0] = a[0];
    else
      s[i] = s[i - 1] + a[i];
  }
  for (i = n - 1; i >= 0; i--) {
    if (i == n - 1)
      d[i] = a[i];
    else
      d[i] = d[i + 1] + a[i];
  }
  ffans = LLONG_MAX;
  for (i = 0; i < n; i++) {
    ans = LLONG_MAX;
    ans2 = LLONG_MAX;
    fans = LLONG_MAX;
    if (i == n - 1 || a[i] != a[i + 1]) {
      if (i >= k - 1) {
        ans = k * a[i] + (i + 1 - k) * (a[i] - 1) - s[i];
      }
      if (n - 1 - i >= k - m[a[i]]) {
        ans2 = d[i + 1] - (n - 1 - i) * (a[i] + 1);
        ans2 += k - m[a[i]];
        if (m[a[i]] >= k) {
          ans2 = 0;
        }
      }
      if (i != 0 && i != n - 1)
        ans = min(ans, (i + 1 - m[a[i]]) * (a[i] - 1) -
                           (i - m[a[i]] >= 0) * s[i - m[a[i]]] + d[i + 1] -
                           (n - 1 - i) * (a[i] + 1) + k - m[a[i]]);
      fans = min(ans, ans2);
      ffans = min(ffans, fans);
    }
  }
  cout << max(0LL, ffans);
}