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() {
  int k, n;
  cin >> n >> k;
  vector<pair<long long int, long long int> > a;
  vector<long long int> tmp(n);
  for (int i = 0; i < n; i++) {
    cin >> tmp[i];
  }
  sort(tmp.begin(), tmp.end());
  long long int ln = 0;
  for (int i = 0; i < n; i++) {
    if (ln != tmp[i]) {
      pair<int, long long int> tt(tmp[i], 1);
      a.push_back(tt);
      ln = tmp[i];
    } else {
      (a[a.size() - 1].second)++;
    }
  }
  vector<long long int> prefcnt(a.size()), prefsum(a.size()), sufsum(a.size()),
      sufcnt(a.size());
  prefcnt[0] = 0;
  prefsum[0] = 0;
  sufsum[a.size() - 1] = 0;
  sufsum[a.size() - 1] = 0;
  long long int as = a[0].first * a[0].second, ac = a[0].second;
  for (int i = 1; i < a.size(); i++) {
    prefcnt[i] = ac;
    prefsum[i] = as;
    as += a[i].first * a[i].second;
    ac += a[i].second;
  }
  as = a[a.size() - 1].first * a[a.size() - 1].second;
  ac = a[a.size() - 1].second;
  for (int i = a.size() - 2; i >= 0; i--) {
    sufcnt[i] = ac;
    sufsum[i] = as;
    as += a[i].first * a[i].second;
    ac += a[i].second;
  }
  long long int cost = 100000000000000;
  for (int i = 0; i < a.size(); i++) {
    if (a[i].second >= k) {
      cost = min((long long int)0, cost);
      break;
    } else {
      long long int c1, c2;
      if (prefcnt[i] + a[i].second < k) {
        c1 = a[i].first * prefcnt[i] - prefsum[i];
        c1 += sufsum[i] - (a[i].first + 1) * sufcnt[i] +
              (k - prefcnt[i] - a[i].second);
      } else {
        c1 = (a[i].first - 1) * prefcnt[i] - prefsum[i] + (k - a[i].second);
      }
      if (sufcnt[i] + a[i].second < k) {
        c2 = sufsum[i] - a[i].first * sufcnt[i];
        c2 += (a[i].first - 1) * prefcnt[i] - prefsum[i] +
              (k - a[i].second - sufcnt[i]);
      } else {
        c2 = sufsum[i] - (a[i].first + 1) * sufcnt[i] + (k - a[i].second);
      }
      c1 = min(c1, c2);
      cost = min(c1, cost);
    }
  }
  cout << cost;
}