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>
const double pi = acos(-1.0);
const double eps = 1e-9;
using namespace std;
int n, k;
void solve() {
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort((a).begin(), (a).end());
  vector<pair<long long, long long>> b;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && a[i] == a[j]) {
      j++;
    }
    b.push_back({a[i], j - i});
    i = j - 1;
  }
  for (auto x : b) {
    if (x.second >= k) {
      cout << 0 << endl;
      return;
    }
  }
  for (auto x : b) {
  }
  int len = b.size();
  vector<long long> prefAns(len), prefKol(len);
  vector<long long> suffAns(len), suffKol(len);
  prefKol[0] = b[0].second;
  prefAns[0] = 0;
  for (int i = 1; i < b.size(); i++) {
    prefAns[i] =
        (b[i].first - b[i - 1].first) * prefKol[i - 1] + prefAns[i - 1];
    prefKol[i] = b[i].second + prefKol[i - 1];
  }
  suffKol[len - 1] = b[len - 1].second;
  suffAns[len - 1] = 0;
  for (int i = b.size() - 2; i >= 0; i--) {
    suffAns[i] =
        (b[i + 1].first - b[i].first) * suffKol[i + 1] + suffAns[i + 1];
    suffKol[i] = b[i].second + suffKol[i + 1];
  }
  long long ans = 1e18;
  for (int i = 0; i < len; i++) {
    long long need = k - b[i].second;
    if (need <= prefKol[i] - b[i].second) {
      long long ansPref = 0;
      if (i > 0) {
        ansPref =
            prefAns[i - 1] + (b[i].first - 1 - b[i - 1].first) * prefKol[i - 1];
        ans = min(ans, ansPref + need);
      }
    }
    if (need <= suffKol[i] - b[i].second) {
      long long ansSuff = 0;
      if (i != len - 1) {
        ansSuff =
            suffAns[i + 1] + (b[i + 1].first - 1 - b[i].first) * suffKol[i + 1];
        ans = min(ans, ansSuff + need);
      }
    }
    if (i > 0 && i != len - 1) {
      if (prefKol[i - 1] < need) {
        long long newNeed = need - prefKol[i - 1];
        assert(newNeed > 0);
        long long del = (b[i + 1].first - 1 - b[i].first) * suffKol[i + 1];
        ans = min(ans, prefAns[i] + suffAns[i + 1] + newNeed + del);
      }
      if (suffKol[i + 1] < need) {
        long long newNeed = need - suffKol[i + 1];
        assert(newNeed > 0);
        long long del = (b[i].first - 1 - b[i - 1].first) * prefKol[i - 1];
        ans = min(ans, prefAns[i - 1] + suffAns[i] + newNeed + del);
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(7);
  int step = 1;
  for (int i = 0; i < step; i++) {
    solve();
  }
}