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;
void setIO(string name) {
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
const int MN = 200 * 1e3 + 5, LN = 17, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int a[MN];
long long ps[MN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) ps[i] = ps[i - 1] + a[i];
  long long mn = LLONG_MAX;
  for (int i = 1, j = 1; i <= n; i = j) {
    for (; j <= n && a[j] == a[i]; j++)
      ;
    long long c = j - i;
    if (c >= k) {
      mn = 0;
      break;
    }
    if (i - 1 >= k - c)
      mn = min(mn, (i - 1) * ((long long)a[i] - 1) - ps[i - 1] + k - c);
    if (n - j + 1 >= k - c)
      mn = min(mn, -(n - j + 1) * ((long long)(a[i] + 1)) + ps[n] - ps[j - 1] +
                       k - c);
    mn = min(mn, (i - 1) * ((long long)a[i] - 1) - ps[i - 1] -
                     (n - j + 1) * ((long long)(a[i] + 1)) + ps[n] - ps[j - 1] +
                     k - c);
  }
  cout << mn << '\n';
  return 0;
}