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 = 200 * 1000 + 20, LG = 18;
int n, a[N], par[N], sz[N], best[1 << LG][2], cnt;
pair<int, int> to[N];
vector<int> vec[1 << LG];
long long ans;
int find_root(int v) { return ~par[v] ? par[v] = find_root(par[v]) : v; }
void merge(int v, int u, int w) {
  v = find_root(v);
  u = find_root(u);
  if (v == u) return;
  if (sz[v] < sz[u]) swap(v, u);
  par[u] = v;
  sz[v] += sz[u];
  ans += w;
  cnt--;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ans -= a[i];
  }
  a[n++] = 0;
  for (int i = 0; i < n; i++) vec[a[i]].push_back(i);
  memset(par, -1, sizeof par);
  fill(sz, sz + n, 1);
  cnt = n;
  for (int mask = 0; mask < (1 << LG); mask++) {
    best[mask][0] = n - 1;
    if (vec[mask].size()) best[mask][0] = vec[mask].back();
    for (int i = 0; i < LG; i++)
      if ((1 << i) & mask)
        if (~best[mask - (1 << i)][0] &&
            a[best[mask - (1 << i)][0]] > a[best[mask][0]])
          best[mask][0] = best[mask - (1 << i)][0];
  }
  while (cnt > 1) {
    for (int mask = 0; mask < (1 << LG); mask++) {
      best[mask][1] = -1;
      for (auto i : vec[mask])
        if (find_root(i) ^ find_root(best[mask][0])) best[mask][1] = i;
      for (int i = 0; i < LG; i++)
        for (int j = 0; j < 2; j++)
          if ((1 << i) & mask)
            if (~best[mask - (1 << i)][j] &&
                find_root(best[mask][0]) != find_root(best[mask - (1 << i)][j]))
              if (best[mask][1] == -1 ||
                  a[best[mask - (1 << i)][j]] > a[best[mask][1]])
                best[mask][1] = best[mask - (1 << i)][j];
    }
    fill(to, to + n, pair<int, int>(-1, -1));
    for (int i = 0; i < n; i++) {
      int mask = (1 << LG) - a[i] - 1;
      for (int j = 0; j < 2; j++)
        if (~best[mask][j])
          if (find_root(best[mask][j]) != find_root(i) &&
              a[i] + a[best[mask][j]] > to[find_root(i)].second)
            to[find_root(i)] = pair<int, int>(find_root(best[mask][j]),
                                              a[i] + a[best[mask][j]]);
    }
    for (int i = 0; i < n; i++)
      if (~to[i].first) merge(i, to[i].first, to[i].second);
  }
  return cout << ans << "\n", 0;
}