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;
const int M = 1000000007, P = 998244353;
const int N = 1000005, E = 262144;
int n, i, j, t, m, k, a[E], f[E], p[E];
long long ans;
int Find(int x) { return (f[x] == -1 ? x : f[x] = Find(f[x])); }
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ++p[a[i]];
    ans -= a[i];
  }
  ++p[0];
  memset(f, -1, sizeof(f));
  for (i = E - 1; i >= 0; --i)
    for (j = i; j; j = (j - 1) & i)
      if (p[j] && p[i ^ j]) {
        int x = Find(j), y = Find(i ^ j);
        if (x != y) {
          f[x] = y;
          ans += 1ll * (p[j] + p[i ^ j] - 1) * i;
          p[j] = p[i ^ j] = 1;
        }
      }
  cout << ans;
}