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 long long INF = 1e18;
const int maxn = 2007;
const int BUBEN = 30;
vector<long long> cnt(1 << 18);
namespace DSU {
const int N = 262144;
int dsu[N];
void init() { fill(dsu, dsu + N, -1); }
int find_set(int u) { return dsu[u] < 0 ? u : dsu[u] = find_set(dsu[u]); }
int union_set(int u, int v) {
  if ((u = find_set(u)) == (v = find_set(v))) {
    return 0;
  }
  if (dsu[u] > dsu[v]) {
    swap(u, v);
  }
  int ans = (dsu[u] == -1 ? cnt[u] : 1) + (dsu[v] == -1 ? cnt[v] : 1) - 1;
  dsu[u] += dsu[v];
  dsu[v] = u;
  return ans;
}
}  // namespace DSU
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n + 1, 0);
  long long sum2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i <= n; i++) {
    sum2 += a[i];
    cnt[a[i]]++;
  }
  DSU::init();
  long long w = 0;
  for (int i = (1 << 18) - 1; i >= 0; i--) {
    int subset = i;
    while (true) {
      int another = i ^ subset;
      if (cnt[subset] != 0 && cnt[another] != 0)
        if (DSU::find_set(subset) != DSU::find_set(another)) {
          long long cnt_edges = (cnt[subset] + cnt[another]) - 1;
          w += i * 1ll * DSU::union_set(subset, another);
        }
      if (subset == 0) break;
      --subset;
      subset &= i;
    }
  }
  cout << w - sum2;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(20);
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}