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;
long long arr[1000000], pref[1000000], suf[1000000];
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  for (long long i = 0; i < n; i++) pref[i + 1] = pref[i] + arr[i];
  long long mini = 1e18;
  for (long long i = 0, j = 0; i < n; i = j) {
    for (; j < n && arr[j] == arr[i]; j++)
      ;
    long long dif = j - i;
    if (dif >= k) {
      cout << 0;
      return 0;
    }
    long long l = ((arr[i] - 1) * i) - pref[i],
              r = pref[n] - pref[j] - ((arr[i] + 1) * (n - j));
    if (i + dif >= k && mini > l + k - dif) mini = l + k - dif;
    if (dif + n - j >= k && mini > r + k - dif) mini = r + k - dif;
    if (mini > l + r + k - dif) mini = l + r + k - dif;
  }
  cout << mini;
  return 0;
}
