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;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n + 1], s = 0;
  map<long long, long long> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    m[a[i]]++;
    s += a[i];
  }
  vector<pair<long long, long long> > v;
  for (auto X : m) {
    v.push_back({X.first, X.second});
  }
  pair<long long, long long> p[v.size() + 2];
  p[0].first = 0;
  p[0].second = 0;
  for (long long i = 0; i < v.size(); i++) {
    p[i + 1].first = p[i].first + ((v[i].first) * (v[i].second));
    p[i + 1].second = p[i].second + v[i].second;
  }
  p[v.size() + 1].first = 0;
  p[v.size() + 1].second = 0;
  long long ans = LLONG_MAX;
  for (long long i = 0; i < v.size(); i++) {
    long long t1 = p[i].first;
    long long t2 = ((v[i].first - 1) * p[i].second);
    t2 -= t1;
    long long t3 = s - p[i + 1].first;
    long long t4 = ((n - p[i + 1].second) * (v[i].first + 1));
    t3 -= t4;
    long long need = LLONG_MAX;
    if (v[i].second >= k) {
      cout << "0\n";
      return;
    }
    long long av = k - v[i].second;
    if (av <= p[i].second) {
      need = min(need, t2 + av);
    }
    if (av <= (n - p[i + 1].second)) {
      need = min(need, t3 + av);
    }
    need = min(need, t2 + t3 + av);
    ans = min(ans, need);
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
}