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 n, k;
long long a[200005];
long long pref[200005];
long long count(long long c) {
  int b = lower_bound(a + 1, a + 1 + n, c) - a,
      e = upper_bound(a + 1, a + 1 + n, c) - a;
  int v = k - (e - b);
  if (v <= 0) return 0;
  long long lc = b * c - pref[b] - (b - 1),
            rc = pref[n] - pref[e - 1] - (n - e + 1) * c - (n - e + 1);
  if ((b - 1) >= v && (n - e + 1) >= v)
    return min(lc + v, rc + v);
  else if ((b - 1) >= v && (n - e + 1) < v)
    return lc + v;
  else if ((b - 1) < v && (n - e + 1) >= v)
    return rc + v;
  else
    return lc + rc + v;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) {
    pref[i] = a[i] + pref[i - 1];
  }
  long long m = 1e17;
  for (int i = 1; i <= n; ++i) {
    long long tm = count(a[i]);
    if (tm < m) m = tm;
  }
  cout << m;
}