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 maxn = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
const int lim = 1 << 18;
int n, x, cnt[maxn], fa[maxn], vis[maxn];
long long ans;
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
void join(long long w, int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  ans += w * ((vis[x] ? 1 : cnt[x]) + (vis[y] ? 1 : cnt[y]) - 1);
  fa[x] = y;
  vis[x] = vis[y] = 1;
}
int main() {
  for (auto i = (1); i <= (lim); ++i) fa[i] = i;
  scanf("%d", &n);
  cnt[0]++;
  for (auto i = (1); i <= (n); ++i) scanf("%d", &x), cnt[x]++, ans -= x;
  for (auto w = (lim); w >= (0); --w)
    for (int u = w; u; u = (u - 1) & w) {
      int v = w ^ u;
      if (cnt[u] && cnt[v]) join(w, u, v);
    }
  printf("%lld\n", ans);
}