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 = 2e5 + 5, MOD = 1e9 + 7, inf = 100000000000000;
long long a[N], pref[N], suff[N];
struct grp {
  long long f = 0, st, en;
};
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  pref[0] = a[0];
  for (long long i = 1; i < n; i++) pref[i] = pref[i - 1] + a[i];
  suff[0] = a[n - 1];
  for (long long i = n - 2, j = 1; i >= 0; i--, j++)
    suff[j] = suff[j - 1] + a[i];
  map<long long, grp> mp;
  mp[a[0]].f++;
  mp[a[0]].st = mp[a[0]].en = 0;
  for (long long i = 1; i < n; i++) {
    mp[a[i]].f++;
    if (a[i] != a[i - 1]) {
      mp[a[i]].st = mp[a[i]].en = i;
      mp[a[i - 1]].en = i - 1;
    }
  }
  if (a[n - 1] == a[n - 2]) mp[a[n - 1]].en = n - 1;
  long long ans = inf;
  for (auto x : mp) {
    if (x.second.f >= k) {
      cout << 0 << "\n";
      return;
    }
    long long idx1, idx2, req = k - x.second.f, temp = 0, val = x.first;
    idx1 = x.second.st - 1;
    idx2 = x.second.en + 1;
    if (req <= idx1 + 1) {
      temp = (idx1 + 1) * (val - 1) - pref[idx1];
      temp += req;
      ans = min(ans, temp);
    } else {
      if (idx1 >= 0) temp = (idx1 + 1) * (val - 1) - pref[idx1];
      if (n - idx2 - 1 >= 0)
        temp += suff[n - idx2 - 1] - (n - idx2) * (val + 1);
      temp += req;
      ans = min(ans, temp);
    }
    if (req <= n - idx2) {
      temp = suff[n - idx2 - 1] - (n - idx2) * (val + 1);
      temp += req;
      ans = min(ans, temp);
    } else {
      if (idx1 >= 0) temp = (idx1 + 1) * (val - 1) - pref[idx1];
      if (n - idx2 - 1 >= 0)
        temp += suff[n - idx2 - 1] - (n - idx2) * (val + 1);
      temp += req;
      ans = min(ans, temp);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
