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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 2e5 + 5;
long long a[N];
map<long long, long long> m;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
    m[a[i]]++;
  }
  long long ans = sum, ps = 0, push_front = 0;
  for (pair<long long, long long> x : m) {
    if (x.second >= k) {
      cout << 0;
      return;
    }
    long long mv = (x.first - 1) * push_front - ps;
    if (push_front + x.second >= k) {
      amin(ans, mv + k - x.second);
    }
    long long mv2 = (sum - ps - x.first * x.second) -
                    (x.first + 1) * (n - push_front - x.second);
    if (n - push_front >= k) {
      amin(ans, mv2 + k - x.second);
    }
    amin(ans, mv + mv2 + k - x.second);
    ps += x.first * x.second;
    push_front += x.second;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}