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 xd[4] = {0, 1, 0, -1}, yd[4] = {1, 0, -1, 0};
template <typename t>
bool ckmin(t& a, const t& b) {
  return a > b ? a = b, true : false;
}
template <typename t>
bool ckmax(t& a, const t& b) {
  return a < b ? a = b, true : false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (auto((i)) = (0); ((i)) < ((n)); ++((i))) cin >> a[i];
  sort((a).begin(), (a).end());
  for (int val = a[0], cnt = 1, i = 1; i < n; ++i) {
    if (cnt == k) {
      cout << 0 << '\n';
      return 0;
    }
    if (a[i] == val)
      cnt++;
    else {
      val = a[i];
      cnt = 1;
    }
  }
  vector<tuple<long long, long long, long long, long long>> prefix(n),
      suffix(n);
  prefix[0] = make_tuple(a[0], 1, 0, 0);
  for (auto(i) = (1); (i) < (n); ++(i)) {
    auto [val, amount, fr, cost] = prefix[i - 1];
    prefix[i] = make_tuple(a[i], amount + 1, a[i] != val ? amount : fr,
                           cost + (a[i] - val) * amount);
  }
  suffix[n - 1] = make_tuple(a[n - 1], 1, 0, 0);
  for (auto((i)) = ((n - 2)); ((i)) > (-1); --((i))) {
    auto [val, amount, fr, cost] = suffix[i + 1];
    suffix[i] = make_tuple(a[i], amount + 1, a[i] != val ? amount : fr,
                           cost + (val - a[i]) * amount);
  }
  long long best = numeric_limits<long long>::max();
  for (int l = 0, r = 0; l < n; ++l) {
    auto [val, amount, fr, cost] = prefix[l];
    if (amount >= k) ckmin(best, cost - min(amount - k, fr));
    while (r + 1 < n && get<0>(suffix[r + 1]) == val &&
           get<1>(suffix[r + 1]) + amount >= k)
      ++r;
    amount += get<1>(suffix[r]);
    ckmin(amount, n);
    fr += get<2>(suffix[r]);
    cost += get<3>(suffix[r]);
    ckmin(best, cost - min(fr, amount - k));
  }
  for (int r = n - 1, l = n - 1; ~r; --r) {
    auto [val, amount, fr, cost] = suffix[r];
    if (amount >= k) ckmin(best, cost - min(amount - k, fr));
    while (l > 0 && get<0>(prefix[l - 1]) == val &&
           get<1>(prefix[l - 1]) + amount >= k)
      --l;
    amount += get<1>(prefix[l]);
    ckmin(amount, n);
    fr += get<2>(prefix[l]);
    cost += get<3>(prefix[l]);
    ckmin(best, cost - min(fr, amount - k));
  }
  cout << best << '\n';
  return 0;
}