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;
namespace myrand {
mt19937 mt(chrono::system_clock::now().time_since_epoch() /
           chrono::microseconds(1));
long long Int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(mt);
}
}  // namespace myrand
using namespace myrand;
auto ss = chrono::system_clock::now();
double remain_time() {
  auto en = chrono::system_clock::now();
  return (chrono::duration_cast<chrono::nanoseconds>(en - ss).count() * 1e-9);
}
long long a[200005];
long long pre_sum[200005];
long long sum[200005];
long long n, k, Left, Right;
long long f(int x) {
  long long total = a[n] * n;
  long long kbe = lower_bound(a + 1, a + 1 + n, x) - a;
  if (a[kbe] >= x) kbe--;
  long long klon = upper_bound(a + 1, a + 1 + n, x) - a;
  long long kkk = (klon - 1) - kbe + (a[kbe] == x);
  long long tt = x * (kbe)-pre_sum[kbe];
  long long t1 = sum[klon] - x * (n - klon + 1);
  kkk = k - kkk;
  klon = n - klon + 1;
  if (kkk <= 0) return 0;
  if (klon >= kkk)
    total = min(total, t1 - (klon - kkk));
  else
    total = min(total, t1 + tt - (kbe - (kkk - klon)));
  if (kbe >= kkk)
    total = min(total, tt - (kbe - kkk));
  else
    total = min(total, (tt + t1 - (klon - (kkk - kbe))));
  return total;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre_sum[i] = 0;
    sum[i] = 0;
  }
  pre_sum[0] = pre_sum[n + 1] = 0;
  sum[0] = sum[n + 1] = 0;
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) pre_sum[i] = pre_sum[i - 1] + a[i];
  for (int i = n; i > 0; i--) sum[i] = sum[i + 1] + a[i];
  long long kq = a[n] * n;
  for (int i = 1; i <= n; i++) {
    long long ttt = f(a[i]);
    kq = min(kq, ttt);
  }
  cout << kq;
  return 0;
}