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 sz = 18;
int f[1 << sz];
struct union_find {
  vector<int> p;
  union_find(int n) : p(n, -1) {}
  int join(int u, int v) {
    if ((u = root(u)) == (v = root(v))) return 0;
    if (p[u] > p[v]) swap(u, v);
    int r = (size(u) == 1 ? f[u] : 1) + (size(v) == 1 ? f[v] : 1) - 1;
    p[u] += p[v];
    p[v] = u;
    return r;
  }
  int root(int u) { return p[u] < 0 ? u : p[u] = root(p[u]); }
  int size(int u) { return -p[root(u)]; }
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  long long ans = 0;
  for (int x; n--;) cin >> x, ++f[x], ans -= x;
  ++f[0];
  union_find uf(1 << sz);
  for (int i = 1 << sz; --i;)
    for (int j = i; 2 * j >= i; --j &= i)
      if (f[i ^ j] && f[j]) ans += (long long)i * uf.join(i ^ j, j);
  cout << ans << '\n';
  return 0;
}