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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, x, z, v, sum, total, ans = 1000000000000000000;
  cin >> n >> k;
  if (k == 1) {
    ans = 0;
  }
  map<long long, long long> m;
  for (long i = 0; i < n; i++) {
    cin >> x;
    m[x]++;
  }
  vector<long long> p, s;
  p.push_back(0);
  s.push_back(0);
  sum = 0;
  total = 0;
  map<long long, long long>::iterator submarine = m.begin(),
                                      helicopter = m.begin();
  helicopter++;
  for (long i = 0; i < m.size() - 1; i++) {
    sum += submarine->second;
    if (sum >= k) {
      ans = min(ans, total - sum + k);
    }
    if (submarine->second >= k) {
      ans = 0;
    }
    total += sum * (helicopter->first - submarine->first);
    p.push_back(total);
    submarine++;
    helicopter++;
  }
  if (submarine->second >= k) {
    ans = 0;
  }
  sum = 0;
  total = 0;
  helicopter--;
  helicopter--;
  for (long i = 0; i < m.size() - 1; i++) {
    sum += submarine->second;
    if (sum >= k) {
      ans = min(ans, total - sum + k);
    }
    total += sum * (submarine->first - helicopter->first);
    s.push_back(total);
    submarine--;
    helicopter--;
  }
  for (long i = 0; i < p.size(); i++) {
    ans = min(ans, p[i] + s[p.size() - 1 - i] - n + k);
  }
  cout << max((long long)0, ans) << endl;
  return 0;
}