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>
const int N = 262144;
using namespace std;
int T, n, m, fa[N], c[N];
int find(int a) { return fa[a] == a ? a : fa[a] = find(fa[a]); }
int merge(int u, int v, int res = 0) {
  u = find(u), v = find(v);
  if (u == v) return 0;
  fa[u] = v;
  return res = c[u] + c[v] - 1, c[u] = c[v] = 1, res;
}
int main() {
  long long ans = 0;
  scanf("%d", &n), c[0] = 1;
  for (int i = 1, w; i <= n; i++) scanf("%d", &w), c[w]++, ans -= w;
  iota(fa, fa + N, 0);
  for (int S = N - 1; S > 0; S--)
    for (int T = S; T > S / 2; T = (T - 1) & S) {
      if (c[T] && c[S ^ T]) ans += 1ll * merge(T, S ^ T) * S;
    }
  printf("%lld\n", ans);
}