Problem: Kuroni isn't good at economics. So he decided to found a new financial pyramid called Antihype. It has the following rules:

  1. You can join the pyramid for free and get 0 coins. 
  2. If you are already a member of Antihype, you can invite your friend who is currently not a member of Antihype, and get a number of coins equal to your age (for each friend you invite). 



n people have heard about Antihype recently, the i-th person's age is a_i. Some of them are friends, but friendship is a weird thing now: the i-th person is a friend of the j-th person if and only if a_i  AND  a_j = 0, where AND denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

Nobody among the n people is a member of Antihype at the moment. They want to cooperate to join and invite each other to Antihype in a way that maximizes their combined gainings. Could you help them? 

Input

The first line contains a single integer n (1≤ n ≤ 2⋅ 10^5) — the number of people.

The second line contains n integers a_1, a_2, ..., a_n (0≤ a_i ≤ 2⋅ 10^5) — the ages of the people.

Output

Output exactly one integer — the maximum possible combined gainings of all n people.

Example

Input


3
1 2 3


Output


2

Note

Only the first and second persons are friends. The second can join Antihype and invite the first one, getting 2 for it.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void cmin(T &a, T b) {
  ((a > b) && (a = b));
}
template <class T>
inline void cmax(T &a, T b) {
  ((a < b) && (a = b));
}
char IO;
template <class T = int>
T rd() {
  T s = 0;
  int f = 0;
  while (!isdigit(IO = getchar())) f |= IO == '-';
  do s = (s << 1) + (s << 3) + (IO ^ '0');
  while (isdigit(IO = getchar()));
  return f ? -s : s;
}
const int N = 1 << 18, INF = 1e9 + 10;
int n, a[N], F[N];
long long ans;
int Find(int x) { return F[x] == x ? x : F[x] = Find(F[x]); }
int main() {
  n = rd();
  for (int i = 1, iend = n; i <= iend; ++i) {
    int x = rd();
    a[x]++, ans -= x;
  }
  a[0]++;
  for (int i = 0, iend = N - 1; i <= iend; ++i) F[i] = i;
  for (int i = N - 1, iend = 1; i >= iend; --i) {
    for (int S = i, T; (S ^ i) <= S; S = (S - 1) & i)
      if (a[S] && a[T = S ^ i] && Find(S) != Find(T)) {
        ans += 1ll * (a[S] + a[T] - 1) * i;
        a[S] = a[T] = 1, F[Find(S)] = Find(T);
      }
  }
  printf("%lld\n", ans);
}