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 K = 18, N = (1 << K) + 2, INF = (int)1e9 + 50;
int n, a[N];
long long sum;
pair<int, int> mx[N][2];
int f[N];
pair<int, int> to[N];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int unite(int u, int v, int val) {
  u = find(u), v = find(v);
  if (u == v)
    return 0;
  else {
    f[u] = v;
    return val;
  }
}
void add(int mask, pair<int, int> p) {
  if (p > mx[mask][0]) {
    if (p.second != mx[mask][0].second) mx[mask][1] = mx[mask][0];
    mx[mask][0] = p;
  } else if (p > mx[mask][1]) {
    if (p.second != mx[mask][0].second) mx[mask][1] = p;
  }
}
void merge(int mask, int from) {
  for (int j = 0; j < 2; j++) add(mask, mx[from][j]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  a[n++] = 0;
  for (int i = 0; i < n; i++) f[i] = i;
  while (true) {
    int cnt = 0;
    fill(to, to + N, pair<int, int>{-INF, -INF});
    for (int i = 0; i < N; i++)
      mx[i][0] = mx[i][1] = pair<int, int>{-INF, -INF};
    for (int i = 0; i < n; i++) {
      add(a[i], {a[i], find(i)});
      cnt += (f[i] == i);
    }
    if (cnt == 1) break;
    for (int i = 0; i < K; i++) {
      for (int j = 0; j < (1 << K); j++)
        if (j & 1 << i) merge(j, j ^ (1 << i));
    }
    for (int i = 0; i < n; i++) {
      int sub = ((1 << K) - 1) ^ a[i];
      int c = find(i);
      pair<int, int> p = (mx[sub][0].second == c ? mx[sub][1] : mx[sub][0]);
      to[c] = max(to[c], pair<int, int>{p.first + a[i], p.second});
    }
    for (int i = 0; i < n; i++) {
      if (f[i] == i) {
        sum += unite(i, to[i].second, to[i].first);
      }
    }
  }
  for (int i = 0; i < n; i++) sum -= a[i];
  cout << sum << '\n';
}