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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
const int N = 1e6 + 4e5 + 66;
const int mod = 1e9 + 7;
const ll inf = 2e9;
const ll INF = 2e18;
const ld EPS = 1e-7;
int a[N];
map<int, int> cnt;
void solve() {
  int n, k;
  cin >> n >> k;
  ll sum = 0, psum = 0, pcn = 0, cn = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    cnt[a[i]]++;
    sum += a[i];
    cn++;
  }
  sort(a + 1, a + 1 + n);
  n = unique(a + 1, a + 1 + n) - a - 1;
  ll ans = INF;
  for (int i = 1; i <= n; ++i) {
    sum -= cnt[a[i]] * 1ll * a[i];
    cn -= cnt[a[i]];
    int l = k - cnt[a[i]];
    if (l <= 0) {
      cout << 0;
      return;
    }
    if (l <= cn) {
      ll s = sum - cn * 1ll * (a[i] + 1) + l;
      ans = min(ans, s);
    }
    if (l <= pcn) {
      ll s = pcn * 1ll * (a[i] - 1) - psum + l;
      ans = min(ans, s);
    }
    if (l <= cn + pcn) {
      ll s = pcn * 1ll * (a[i] - 1) - psum + sum - cn * 1ll * (a[i] + 1) + l;
      ans = min(ans, s);
    }
    psum += cnt[a[i]] * 1ll * a[i];
    pcn += cnt[a[i]];
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
}