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;
map<long long, long long> mp;
map<long long, long long>::iterator it;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a = 0, b = 0, c, d, e, f = 0, l, g, m, n, k, i, j, t, p, q, sum = 0,
            ssum, pp;
  cin >> n >> k;
  f = 1234567891253456;
  for (i = 0; i < n; i++) {
    cin >> a;
    sum += a;
    mp[a]++;
    if (mp[a] >= k) f = 0;
  }
  a = 0;
  ssum = 0;
  it = mp.begin();
  while (it != mp.end()) {
    d = k - it->second;
    if (d <= 0) break;
    if (a >= d) {
      p = it->first - 1;
      p *= a;
      p -= ssum;
      p += d;
      f = min(f, p);
    }
    if (n - a >= k) {
      p = it->first + 1;
      q = n - a - it->second;
      p *= q;
      p = sum - ssum - it->second * it->first - p;
      p += d;
      f = min(f, p);
    }
    p = it->first - 1;
    p *= a;
    p -= ssum;
    pp = it->first + 1;
    q = n - a - it->second;
    pp *= q;
    pp = sum - ssum - it->second * it->first - pp;
    p += pp + d;
    f = min(f, p);
    a += it->second;
    ssum += it->first * it->second;
    it++;
  }
  cout << f << '\n';
  return 0;
}