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;
const int N = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  map<int, long long> fr;
  long long cnt = 0, prefSum = 0, sum = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    sum += a;
    fr[a]++;
  }
  long long ans = 1e18;
  for (auto it : fr) {
    if (it.second >= k) {
      ans = 0;
      break;
    }
    {
      long long rem = k - it.second;
      long long cur = (it.first - 1) * cnt - prefSum;
      cur += min(rem, cnt);
      if (cnt >= rem)
        ans = min(ans, cur), rem = 0;
      else
        rem -= cnt;
      cur += (sum - prefSum - it.first * 1LL * it.second) -
             (it.first + 1) * (n - cnt - it.second);
      cur += rem;
      ans = min(ans, cur);
    }
    {
      long long rem = k - it.second;
      long long cur = (sum - prefSum - it.first * 1LL * it.second) -
                      (it.first + 1) * (n - cnt - it.second);
      cur += min(rem, (n - cnt - it.second));
      if ((n - cnt - it.second) >= rem)
        ans = min(ans, cur), rem = 0;
      else
        rem -= (n - cnt - it.second);
      cur += (it.first - 1) * cnt - prefSum;
      ;
      cur += rem;
      ans = min(ans, cur);
    }
    prefSum += it.second * 1LL * it.first;
    cnt += it.second;
  }
  cout << ans << endl;
  return 0;
}