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 INF = 1000000000;
const pair<int, int> X = make_pair(-INF, 0);
int f[270500];
int find(int u) { return f[u] == u ? u : f[u] = find(f[u]); }
bool merge(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return 0;
  f[u] = v;
  return 1;
}
struct Data {
  pair<int, int> x, y;
  void upd(const pair<int, int> &B) {
    if (B.first > x.first) {
      if (B.second != x.second) y = x;
      x = B;
    } else if (B.second != x.second)
      y = max(y, B);
  }
  void upd(const Data &B) {
    upd(B.x);
    upd(B.y);
  }
} o[270500];
pair<int, int> to[270500];
int n, a[270500];
long long ans;
int main() {
  scanf("%d", &n);
  int lim = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    while (lim <= a[i]) lim <<= 1;
  }
  a[++n] = 0;
  for (int u = 1; u <= n; u++) f[u] = u;
  for (int cntl = 0; cntl < n - 1;) {
    for (int s = 0; s < lim; s++) o[s] = (Data){X, X};
    for (int u = 1; u <= n; u++) {
      to[u] = X;
      find(u);
      o[a[u]].upd(make_pair(a[u], f[u]));
    }
    for (int l = 1; l < lim; l <<= 1)
      for (int p = 0; p < lim; p += l + l)
        for (int k = 0; k < l; k++) o[p | k | l].upd(o[p | k]);
    for (int u = 1; u <= n; u++) {
      int ts = (lim - 1) ^ a[u];
      pair<int, int> sav = (o[ts].x.second == f[u]) ? o[ts].y : o[ts].x;
      sav.first += a[u];
      to[f[u]] = max(to[f[u]], sav);
    }
    for (int u = 1; u <= n; u++)
      if (f[u] == u && to[u].first >= 0 && merge(u, to[u].second)) {
        ans += to[u].first;
        cntl++;
      }
  }
  long long sum = 0;
  for (int u = 1; u <= n; u++) sum += a[u];
  printf("%lld", ans - sum);
  return 0;
}