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 LG = 18;
int n, par[(1 << LG) + 20];
long long cnt[(1 << LG) + 20], ans;
int get_par(int v) { return (v == par[v] ? v : par[v] = get_par(par[v])); }
void Merge(int x, int y) {
  int v = get_par(x), u = get_par(y);
  if (u == v) return;
  par[v] = u;
  ans += 1LL * (cnt[x] + cnt[y] - 1) * (x + y);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1, x; i <= n; i++) {
    cin >> x;
    cnt[x]++;
    ans -= x;
  }
  for (int i = 1; i < (1 << LG); i++) par[i] = i;
  cnt[0]++;
  for (int mask = (1 << LG) - 1; ~mask; mask--) {
    for (int i = mask; i; i = (i - 1) & mask) {
      int v = i, u = i ^ mask;
      if (!cnt[u] || !cnt[v]) continue;
      Merge(v, u);
      cnt[v] = 1, cnt[u] = 1;
    }
  }
  cout << ans << endl;
  return 0;
}