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;
template <class T>
int chkmax(T& a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int chkmin(T& a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class iterator>
void output(iterator begin, iterator end, ostream& out = cerr) {
  while (begin != end) {
    out << (*begin) << " ";
    begin++;
  }
  out << endl;
}
template <class T>
void output(T x, ostream& out = cerr) {
  output(x.begin(), x.end(), out);
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mx = (1 << 18);
int p[mx], rnk[mx];
void init() { iota(p, p + mx, 0); }
int get_root(int v) { return (p[v] == v ? v : p[v] = get_root(p[v])); }
int join(int u, int v) {
  u = get_root(u);
  v = get_root(v);
  if (u != v) {
    if (rnk[u] > rnk[v]) {
      swap(u, v);
    }
    p[u] = v;
    if (rnk[u] == rnk[v]) {
      rnk[v]++;
    }
    return 1;
  } else {
    return 0;
  }
}
long long ans = 0;
int n, cnt[mx], conn[mx];
signed main() {
  fast_io();
  init();
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int val;
    cin >> val;
    ans -= val;
    cnt[val]++;
  }
  cnt[0]++;
  for (int e = mx - 1; e >= 0; --e) {
    for (int i = e; true; i = ((i - 1) & e)) {
      if (cnt[i] && cnt[e ^ i] && join(i, e ^ i)) {
        ans += 1LL * (cnt[i] + cnt[e ^ i] - 1) * e;
        cnt[i] = 1;
        cnt[e ^ i] = 1;
      }
      if (i == 0) {
        break;
      }
    }
  }
  cout << ans << endl;
}