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;
using ll = long long;
const int inf = 1e9 + 10;
const ll inf_ll = 1e18 + 10;
const int N = 1 << 18;
int f[N], v[N], c[N], p[N], sz[N], val[N];
int rep(int x) {
  while (x != p[x]) x = p[x] = p[p[x]];
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ll n;
  cin >> n;
  iota(p, p + N, 0);
  iota(val, val + N, 0);
  fill(sz, sz + N, 1);
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    ll x;
    cin >> x;
    ans -= x, f[x]++;
  }
  for (int i = N - 1; i > 0; i--) {
    for (int j = i; j > i - j; j = (j - 1) & i) {
      if (!f[j] || !f[i - j]) continue;
      int x = rep(i - j), y = rep(j);
      if (x != y && min(val[x], val[y]) < i) {
        ll z = 1;
        if (!v[j]) v[j] = 1, z += f[j] - 1;
        if (!v[i - j]) v[i - j] = 1, z += f[i - j] - 1;
        ans += z * i;
        if (sz[x] < sz[y]) swap(x, y);
        p[y] = x, sz[x] += sz[y];
        val[x] = max(val[x], val[y]);
      }
    }
  }
  for (int i = 0; i < N; i++)
    if (f[i]) (c[rep(i)] = max(c[rep(i)], i));
  for (int i = 0; i < N; i++)
    if (f[i]) {
      if (sz[rep(i)] == 1)
        ans += ll(f[i]) * i;
      else if (c[i] > 0)
        ans += c[i];
    }
  cout << ans << "\n";
}