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 N = 1 << 18;
int n, u, cnt[N];
long long ans = 0;
struct dsu {
  int dsu[N];
  void init() { fill(dsu, dsu + N, -1); }
  int trace(int u) { return dsu[u] < 0 ? u : dsu[u] = trace(dsu[u]); }
  int connect(int u, int v) {
    if ((u = trace(u)) == (v = trace(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;
  }
} dsu;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> u;
    cnt[u]++;
    ans -= u;
  }
  cnt[0]++;
  dsu.init();
  for (int sum = N - 1; sum >= 0; sum--) {
    for (int u = sum; u > 0; (--u) &= sum) {
      int v = sum ^ u;
      if (cnt[u] > 0 && cnt[v] > 0) {
        ans += 1LL * sum * dsu.connect(u, v);
      }
    }
  }
  cout << ans;
}