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;
int a[200001];
int parent[200001];
int find(int n) {
  if (parent[n] != n) parent[n] = find(parent[n]);
  return parent[n];
}
pair<int, int> best[1 << 18][2];
pair<int, int> pp[200001];
int main() {
  int i;
  int n;
  long long int ans = 0;
  scanf("%d", &n), n++;
  for (i = 1; i < n; i++) scanf("%d", &a[i]), ans -= a[i], parent[i] = i;
  int j, c = n;
  while (c > 1) {
    for (i = 0; i < (1 << 18); i++) best[i][0] = best[i][1] = make_pair(-1, -1);
    for (i = 0; i < n; i++) {
      pair<int, int> p = make_pair(a[i], find(i));
      if (p.first >= best[a[i]][0].first) {
        if (p.second == best[a[i]][0].second)
          best[a[i]][0] = p;
        else
          best[a[i]][1] = best[a[i]][0], best[a[i]][0] = p;
      } else if (p.first >= best[a[i]][1].first) {
        if (p.second != best[a[i]][0].second) best[a[i]][1] = p;
      }
    }
    for (i = 0; i < 18; i++) {
      for (j = 0; j < (1 << 18); j++) {
        if (j & (1 << i)) {
          pair<int, int> p = best[j ^ (1 << i)][0];
          if (p.first >= best[j][0].first) {
            if (p.second == best[j][0].second)
              best[j][0] = p;
            else
              best[j][1] = best[j][0], best[j][0] = p;
          } else if (p.first >= best[j][1].first) {
            if (p.second != best[j][0].second) best[j][1] = p;
          }
          p = best[j ^ (1 << i)][1];
          if (p.first >= best[j][0].first) {
            if (p.second == best[j][0].second)
              best[j][0] = p;
            else
              best[j][1] = best[j][0], best[j][0] = p;
          } else if (p.first >= best[j][1].first) {
            if (p.second != best[j][0].second) best[j][1] = p;
          }
        }
      }
    }
    for (i = 0; i < n; i++) pp[i] = make_pair(-1, -1);
    vector<int> r;
    for (i = 0; i < n; i++) {
      pair<int, int> p;
      if (best[((1 << 18) - 1) ^ a[i]][0].second == find(i))
        p = best[((1 << 18) - 1) ^ a[i]][1];
      else
        p = best[((1 << 18) - 1) ^ a[i]][0];
      if (p.second != -1)
        pp[find(i)] = max(pp[find(i)], make_pair(p.first + a[i], p.second));
      if (i == find(i)) r.push_back(i);
    }
    for (i = 0; i < r.size(); i++) {
      if (find(pp[r[i]].second) != find(r[i])) {
        ans += pp[r[i]].first;
        parent[find(pp[r[i]].second)] = find(r[i]), c--;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}