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 N = 2e5 + 5;
pair<long long, long long> pre[N];
pair<long long, long long> suf[N];
int n, k, t;
unordered_map<int, int> ma;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  bool ok = false;
  for (int i = 0; i < n; ++i) {
    cin >> t;
    ++ma[t];
    if (ma[t] >= k) ok = true;
  }
  if (ok) {
    cout << 0 << endl;
    exit(0);
  }
  vector<pair<long long, long long> > v(ma.begin(), ma.end());
  sort(v.begin(), v.end());
  pair<long long, long long> p = {v[0].second, 0};
  pre[0] = p;
  for (int i = 1; i < v.size(); ++i) {
    p.second += p.first * (v[i].first - v[i - 1].first);
    p.first += v[i].second;
    pre[i] = p;
  }
  p = {v.back().second, 0};
  suf[v.size() - 1] = p;
  for (int i = v.size() - 2; i >= 0; --i) {
    p.second += p.first * (v[i + 1].first - v[i].first);
    p.first += v[i].second;
    suf[i] = p;
  }
  long long res = 0x3f3f3f3f3f3f3f3f;
  for (int i = 0; i < v.size(); ++i) {
    auto [lcnt, lmov] = pre[i];
    auto [rcnt, rmov] = suf[i];
    if (lcnt >= k) {
      res = min(res, lmov - (lcnt - k));
    }
    if (rcnt >= k) {
      res = min(res, rmov - (rcnt - k));
    }
    if (lcnt < k && rcnt < k) {
      res = min(res, lmov + rmov - (n - k));
    }
  }
  cout << res << endl;
  return 0;
}