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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4,avx2,abm,fma,tune=native")
using namespace std;
const char nl = '\n';
const int N = 1 << 18;
int cnt[N];
int dsu[N], sz[N];
int find(int i) {
  if (dsu[i] == -1) return i;
  return dsu[i] = find(dsu[i]);
}
int link(int i, int j) {
  i = find(i);
  j = find(j);
  if (i == j) return 0;
  if (sz[i] > sz[j]) swap(i, j);
  int edges = cnt[i] + cnt[j] - 1;
  cnt[i] = cnt[j] = 1;
  dsu[i] = j;
  sz[j] += i;
  return edges;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  memset(dsu, -1, sizeof dsu);
  fill(sz, sz + N, 1);
  int n;
  cin >> n;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    int v;
    cin >> v;
    cnt[v]++;
    sum += v;
  }
  cnt[0]++;
  long long ans = 0;
  for (int cost = N - 1; cost >= 0; cost--) {
    for (int u = cost; u > 0; u = (u - 1) & cost) {
      int v = cost ^ u;
      if (cnt[u] && cnt[v]) {
        ans += (long long)cost * link(u, v);
      }
    }
  }
  cout << ans - sum << nl;
  return 0;
}