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;
long long P = 1e9 + 7;
long long P1 = 998244353;
const int sqr = 500;
const int arrn = 4e6 + 5;
const int arrn2 = 6e3 + 5;
void solve() {
  int n, k;
  cin >> n >> k;
  map<long long, int> cnt;
  long long tot = 0;
  for (int i = 0; i < n; i++) {
    long long k;
    cin >> k;
    tot += k;
    cnt[k]++;
  }
  int lft = 0;
  long long lsum = 0;
  long long ans = LLONG_MAX;
  for (auto i : cnt) {
    int rem = k - i.second;
    int rig = n - i.second - lft;
    long long rsum = tot - lsum - (i.first * i.second);
    if (rem <= 0) {
      ans = 0;
      break;
    }
    if (rem <= lft) {
      long long oper = (i.first - 1) * lft - lsum;
      oper += rem;
      ans = min(ans, oper);
    }
    if (rem <= rig) {
      long long oper = rsum - (i.first + 1) * rig;
      oper += rem;
      ans = min(ans, oper);
    }
    long long oper =
        (rsum - (i.first + 1) * rig) + ((i.first - 1) * lft - lsum);
    oper += rem;
    ans = min(ans, oper);
    lft += i.second;
    lsum += i.first * i.second;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}