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;
  scanf("%d%d", &n, &k);
  long long a[n];
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  long long sum[n + 1];
  sum[0] = 0;
  sort(a, a + n);
  for (int i = 1; i < n + 1; i++) sum[i] = sum[i - 1] + a[i - 1];
  long long mini, x = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      x = 0;
    }
    x++;
    if (x >= k) {
      cout << 0;
      return 0;
    }
  }
  int ind1, ind2;
  ind1 = lower_bound(a, a + n, a[k - 1]) - a - 1;
  ind2 = upper_bound(a, a + n, a[k - 1]) - a - 1;
  x = (a[k - 1] - 1) * (ind1 + 1) - sum[ind1 + 1] + k - ind2 + ind1;
  mini = x;
  ind1 = upper_bound(a, a + n, a[n - k]) - a;
  ind2 = lower_bound(a, a + n, a[n - k]) - a;
  x = sum[n] - sum[ind1] - (a[n - k] + 1) * (n - ind1) + k - ind1 + ind2;
  mini = min(x, mini);
  if (k > n / 2) {
    ind1 = lower_bound(a, a + n, a[n / 2]) - a;
    ind2 = upper_bound(a, a + n, a[n / 2]) - a;
    x = a[n / 2] * ind1 - sum[ind1] + sum[n] - sum[ind2] -
        (a[n / 2] + 1) * (n - ind2);
    if (k > ind2) x += k - ind2;
    mini = min(x, mini);
  }
  cout << mini;
}