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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
T gcd(T a, T b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < (int)(n); i++) cin >> a[i];
  sort(a.begin(), a.end());
  vector<pair<long long, int>> b;
  int cnt = 1;
  for (int i = (int)(1); i < (int)(n); i++) {
    if (a[i] == a[i - 1])
      cnt++;
    else {
      b.push_back(make_pair(a[i - 1], cnt));
      cnt = 1;
    }
  }
  b.push_back(make_pair(a[n - 1], cnt));
  long long sum = 0, num = 0;
  long long tot = 0, num1 = n;
  long long ans = 1e18;
  for (int i = 0; i < (int)(b.size()); i++)
    tot += (b[i].first * (long long)(b[i].second));
  for (int i = 0; i < (int)(b.size()); i++) {
    tot -= (b[i].first * (long long)(b[i].second));
    num1 -= b[i].second;
    int cur = b[i].second;
    if (cur >= k) {
      ans = 0;
      break;
    }
    int need = k - cur;
    long long lw = ((b[i].first - 1) * num) - sum;
    long long up = tot - ((b[i].first + 1) * num1);
    int f = 0;
    if (num >= need) f |= 1;
    if (num1 >= need) f |= 2;
    if (f == 3) ans = min(ans, min(up, lw) + (long long)(need));
    if (f == 2) ans = min(ans, up + (long long)(need));
    if (f == 1) ans = min(ans, lw + (long long)(need));
    if (!f && (num + num1) >= need) ans = min(ans, lw + up + (long long)(need));
    num += (long long)(b[i].second);
    sum += (b[i].first * (long long)(b[i].second));
  }
  cout << ans << '\n';
  return 0;
}