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;
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  map<int, int> fre;
  vector<long long> a(n), f(n, 0), r(n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    fre[a[i]]++;
  }
  sort(a.begin(), a.end());
  long long sum = a[0];
  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1])
      f[i] = f[i - 1];
    else
      f[i] = (a[i] - 1) * i - sum;
    sum += a[i];
    2;
  }
  sum = a[n - 1];
  for (int i = n - 2; i >= 0; --i) {
    if (a[i] == a[i + 1])
      r[i] = r[i + 1];
    else
      r[i] = sum - (a[i] + 1) * (n - i - 1);
    sum += a[i];
  }
  for (int i = 0; i < n; ++i) {
    if (fre[a[i]] >= k) {
      cout << 0 << '\n';
      return 0;
    }
  }
  2;
  2;
  2;
  long long ans = LLONG_MAX;
  for (int i = 0; i < n; ++i) {
    if (i >= k - 1) ans = min(ans, f[i] + max(0, k - fre[a[i]]));
    if (i <= n - k)
      ans = min(ans, r[i] + max(0, k - fre[a[i]]));
    else
      ans = min(ans, f[i] + r[i] + max(0, k - fre[a[i]]));
  }
  cout << ans << '\n';
}
