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")
using namespace std;
const long long N = 1e6 + 10;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long eps = 1e-9;
const char spc = 29;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
long long n, k;
long long a[N];
long long calc(long long first) {
  long long grcnt = 0;
  long long lscnt = 0;
  long long grops = 0;
  long long lsops = 0;
  long long lft = k;
  for (long long i = 0; i < n; ++i)
    if (a[i] == first)
      --lft;
    else if (a[i] > first) {
      ++grcnt;
      grops += a[i] - first - 1;
    } else {
      ++lscnt;
      lsops += first - a[i] - 1;
    }
  long long res = inf;
  if (grcnt >= lft) res = min(res, grops + lft);
  if (lscnt >= lft) res = min(res, lsops + lft);
  res = min(res, lsops + grops + lft);
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  map<long long, long long> cnts;
  long long sum = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    ++cnts[a[i]];
    sum += a[i];
  }
  long long lscnt = 0;
  long long lssum = 0;
  long long res = inf;
  for (auto &p : cnts) {
    long long curcnt = p.second;
    long long lft = k - curcnt;
    long long grsum = sum - lssum - (curcnt * p.first);
    long long grcnt = n - lscnt - curcnt;
    if (lft <= 0) {
      res = 0;
      break;
    }
    if (grcnt >= lft) res = min(res, grsum - grcnt * (p.first + 1) + lft);
    if (lscnt >= lft) res = min(res, lscnt * (p.first - 1) - lssum + lft);
    res = min(res, -lssum + grsum + lscnt * (p.first - 1) -
                       grcnt * (p.first + 1) + lft);
    lscnt += curcnt;
    lssum += p.first * p.second;
  }
  cout << res << "\n";
}