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 INF = 1e18;
const int N = 2e5 + 7;
int arr[N];
long long sum[N];
map<int, int> mp;
int main() {
  int n, k, num = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++)
    scanf("%d", &arr[i]), mp[arr[i]]++, num = max(num, mp[arr[i]]);
  if (num >= k) {
    cout << 0 << endl;
    return 0;
  }
  sort(arr + 1, arr + n + 1);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + arr[i];
  long long ans = INF;
  for (int i = 1; i <= n; i++) {
    if (i >= k) ans = min(ans, (long long)arr[i] * (i)-sum[i] - (i - k));
    if (n - i + 1 >= k)
      ans = min(
          ans, sum[n] - sum[i] - (long long)arr[i] * (n - i) - (n - i + 1 - k));
    ans = min(ans, (long long)arr[i] * (i)-sum[i] + sum[n] - sum[i] -
                       (long long)arr[i] * (n - i) - (n - k));
  }
  cout << ans << endl;
  return 0;
}
