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 MAX_N = 200005;
const int MAX_V = 1 << 18;
int l[MAX_N];
int n;
int par[MAX_V];
int sz[MAX_V];
int find(int x) { return par[x] == x ? x : (par[x] = find(par[x])); }
long long merge(int x, int y) {
  long long ret = sz[x] + (long long)sz[y] - 1;
  int ox = x;
  int oy = y;
  x = find(x);
  y = find(y);
  if (x == y) return 0;
  sz[ox] = 1;
  sz[oy] = 1;
  if (rand() & 1)
    par[x] = y;
  else
    par[y] = x;
  return ret;
}
void solve() {
  cin >> n;
  sz[0]++;
  for (int i = 1; i <= n; i++) {
    cin >> l[i];
    sz[l[i]]++;
  }
  for (int i = 1; i < MAX_V; i++) par[i] = i;
  long long ret = 0;
  for (int i = MAX_V - 1; i > 0; i--) {
    int mask = (MAX_V - 1) & i;
    while (mask > 0) {
      int othermask = mask ^ i;
      if (othermask > mask) break;
      if (sz[mask] && sz[othermask]) {
        ret += merge(mask, othermask) * i;
      }
      mask = (mask - 1) & i;
    }
  }
  for (int i = 1; i <= n; i++) ret -= l[i];
  cout << ret << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}