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;
using LL = long long;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  map<int, int> cnt;
  for (int i = 0, x; i < n; ++i) {
    cin >> x;
    cnt[x] += 1;
  }
  vector<pair<int, int>> a;
  for (auto p : cnt) {
    a.push_back(p);
  }
  vector<pair<LL, int>> cl(a.size()), cr(a.size());
  LL sum = 0, cc = 0;
  for (int i = 0; i < a.size(); ++i) {
    auto [x, y] = a[i];
    cl[i] = {cc * (x - 1) - sum, cc};
    sum += 1LL * x * y;
    cc += y;
  }
  sum = 0;
  cc = 0;
  for (int i = (int)a.size() - 1; i >= 0; --i) {
    auto [x, y] = a[i];
    cr[i] = {sum - cc * (x + 1), cc};
    sum += 1LL * x * y;
    cc += y;
  }
  LL ans = numeric_limits<LL>::max();
  for (int i = 0; i < (int)a.size(); ++i) {
    if (a[i].second >= k) {
      ans = 0;
      break;
    }
    if (a[i].second + cl[i].second >= k) {
      ans = min(ans, cl[i].first + k - a[i].second);
    }
    if (a[i].second + cr[i].second >= k) {
      ans = min(ans, cr[i].first + k - a[i].second);
    }
    ans = min(ans, cl[i].first + cr[i].first + k - a[i].second);
  }
  cout << ans << '\n';
}