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
#include <bits/stdc++.h>
using namespace std;
const long long N = 100005;
long long n, k;
vector<long long> vect, pre, suf;
long long ans = 1000000000000000007LL;
void play(long long u) {
  long long x = vect[u];
  long long ct = upper_bound(vect.begin(), vect.end(), x) -
                 lower_bound(vect.begin(), vect.end(), x);
  long long lp =
      lower_bound(vect.begin(), vect.end(), x - 1) - vect.begin() - 1;
  long long lreq = 0;
  if (lp >= 0) {
    lreq = (lp + 1) * (x - 1) - pre[lp];
  }
  long long rp = upper_bound(vect.begin(), vect.end(), x + 1) - vect.begin();
  long long rreq = 0;
  if (rp < n) {
    rreq = suf[rp] - (n - rp) * (x + 1);
  }
  long long ls = lower_bound(vect.begin(), vect.end(), x) - vect.begin();
  long long rs = n - (upper_bound(vect.begin(), vect.end(), x) - vect.begin());
  if (ls + ct >= k) {
    ans = min(ans, lreq + (k - ct));
  }
  if (rs + ct >= k) {
    ans = min(ans, rreq + (k - ct));
  }
  ans = min(ans, lreq + rreq + (k - ct));
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  vect.resize(n);
  pre.resize(n);
  suf.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> vect[i];
  }
  sort(vect.begin(), vect.end());
  pre[0] = vect[0];
  for (long long i = 1; i < n; i++) {
    pre[i] = pre[i - 1] + vect[i];
  }
  suf[n - 1] = vect[n - 1];
  for (long long i = n - 2; i >= 0; i--) {
    suf[i] = suf[i + 1] + vect[i];
  }
  long long ct = 1;
  long long mx = 1;
  long long cur = vect[0];
  for (long long i = 1; i < (long long)vect.size(); i++) {
    if (vect[i] == cur) {
      ct++;
      mx = max(mx, ct);
    } else {
      cur = vect[i];
      ct = 1;
    }
  }
  mx = max(mx, ct);
  if (mx >= k) {
    cout << 0;
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    play(i);
  }
  cout << ans;
  return 0;
}
