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 N = 1 << 18;
int fa[N];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
long long ans = 0;
int n, cnt[N], vis[N];
int getcnt(int x) {
  if (vis[x]) return 1;
  vis[x] = 1;
  return cnt[x];
}
void connect(long long w, int x, int p) {
  x = find(x);
  p = find(p);
  if (x != p) fa[x] = p, ans += w * (getcnt(x) + getcnt(p) - 1);
}
signed main() {
  cin >> n;
  for (int i = (1), _ = (N); i < _; i++) fa[i] = i;
  for (int i = (0), _ = (n); i < _; i++) {
    int x;
    cin >> x;
    cnt[x]++;
    ans -= x;
  }
  cnt[0]++;
  for (int s = (N)-1, _ = (0); s >= _; s--)
    for (int x = s; x; x = (x - 1) & s) {
      int p = s ^ x;
      if (cnt[x] && cnt[p]) connect(s, x, p);
    }
  cout << ans << endl;
  return 0;
}