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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, i, costL = 0, costR = 0, x, mx = 0;
  vector<long long> v;
  map<long long, long long> cnt;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back(x);
    cnt[x]++;
    mx = max(mx, cnt[x]);
  }
  if (mx >= k) {
    cout << 0 << "\n";
    return 0;
  }
  sort(v.begin(), v.end());
  for (i = 0; i < k; i++) {
    costL += v[k - 1] - v[i];
  }
  for (i = k; i < n; i++) {
    if (v[k - 1] == v[i])
      costL--;
    else
      break;
  }
  for (i = n - k; i < n; i++) {
    costR += v[i] - v[n - k];
  }
  for (i = k; i < n; i++) {
    if (v[n - k] == v[n - i - 1])
      costR--;
    else
      break;
  }
  long long total = 0LL;
  for (i = 0; i + i < n; i++) {
    total += v[n - i - 1] - v[i];
  }
  total -= (n - k);
  long long ans = min({total, costL, costR});
  cout << max(0LL, ans) << "\n";
  return 0;
}