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;
long long int binary_exp(long long int a, long long int n) {
  long long int i = 1;
  while (n > 0) {
    if (n % 2 == 1) i = (i * a) % 998244353;
    a = (a * a) % 998244353;
    n /= 2;
  }
  return i;
}
long long int IsPrime(long long int n) {
  if (n == 0 || n == 1) return n;
  if (n % 2 == 0) return (long long int)2;
  if (n % 3 == 0)
    return (long long int)3;
  else {
    for (long long int i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
void solve() {
  long long int n, i, j, k, a, b, pre = 0, tot = 0, ans = LLONG_MAX, req,
                                  pre_sum = 0, temp1, temp2;
  cin >> n >> k;
  map<long long int, long long int> freq;
  for (i = 0; i < n; i++) {
    cin >> a;
    tot += a;
    freq[a]++;
  }
  map<long long int, long long int>::iterator itr, itr1, itr2;
  for (itr = freq.begin(); itr != freq.end(); itr++) {
    a = b = 1e10;
    tot -= (itr->first * itr->second);
    if (itr->second >= k) {
      cout << 0;
      return;
    }
    req = k - itr->second;
    if (pre > 0) a = pre * (itr->first - 1) - pre_sum;
    if (n - pre - itr->second > 0)
      b = tot - (n - pre - itr->second) * (itr->first + 1);
    temp1 = min(pre, req) + a;
    req -= min(pre, req);
    if (req > 0) temp1 += (min(req, n - pre - itr->second) + b);
    req = k - itr->second;
    temp2 = min(n - pre - itr->second, req) + b;
    req -= min(n - pre - itr->second, req);
    if (req > 0) temp2 += (min(req, pre) + a);
    ans = min(ans, min(temp1, temp2));
    pre += itr->second;
    pre_sum += (itr->first * itr->second);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  int t = 1;
  while (t--) solve();
}