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 nmax = (1 << 18) + 42;
int n, inp[nmax];
int parent[nmax];
vector<int> seen[nmax];
int where[nmax], coeff[nmax];
long long outp;
int main() {
  scanf("%i", &n);
  for (int i = 1; i <= n; i++) scanf("%i", &inp[i]);
  n++;
  inp[n] = 0;
  for (int i = 1; i <= n; i++) parent[i] = i;
  sort(inp + 1, inp + n + 1);
  for (int i = 1; i <= n; i++) {
    where[inp[i]] = i;
    coeff[inp[i]]++;
  }
  for (int i = 0; i <= n; i++) seen[i].push_back(i);
  outp = 0;
  for (int i = 1; i <= n; i++) outp = outp - inp[i];
  for (int sum = (1 << 18) - 1; sum >= 0; sum--) {
    for (int i = sum; i; i = (i - 1) & sum) {
      int j = sum - i;
      if (where[i] == 0 || where[j] == 0) continue;
      if (parent[where[i]] == parent[where[j]]) continue;
      outp += (1LL) * (i + j) * (coeff[i] + coeff[j] - 1);
      coeff[i] = 1;
      coeff[j] = 1;
      int u = parent[where[i]], v = parent[where[j]];
      if (seen[u].size() < seen[v].size()) swap(u, v);
      for (auto w : seen[v]) {
        parent[w] = u;
        seen[u].push_back(w);
      }
    }
  }
  printf("%lld\n", outp);
  return 0;
}