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() {
  int n, k;
  cin >> n >> k;
  ;
  vector<long long> arr(n + 1, 0);
  for (int i = 1; i <= n; i++) cin >> arr[i];
  ;
  sort(arr.begin(), arr.end());
  vector<long long> prefix(2e5 + 5, 0), suffix(2e5 + 5, 0);
  for (int i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] + arr[i];
  }
  for (int i = n; i >= 1; i--) {
    suffix[i] = suffix[i + 1] + arr[i];
  }
  long long ans = 1e15;
  for (int i = 1; i <= n; i++) {
    int right = i + 1;
    while (right < n && arr[i] == arr[right]) right++;
    int have = right - i;
    if (have >= k) {
      cout << 0 << endl;
      return 0;
    }
    int need = k - have;
    long long tmp = (arr[i] - 1) * (i - 1) - prefix[i - 1] + suffix[right] -
                    (arr[i] + 1) * (n - right + 1) + need;
    ans = min(ans, tmp);
    if (need < i) {
      ans = min(ans, (arr[i] - 1) * (i - 1) - prefix[i - 1] + need);
    }
    if (n - right + 1 >= need) {
      ans = min(ans, suffix[right] - (arr[i] + 1) * (n - right + 1) + need);
    }
    i = right - 1;
  }
  cout << ans << endl;
}