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 read() {
  int a = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return a;
}
const int _ = 1 << 18;
int fa[_], cnt[_], sz[_], N;
long long sum = 0;
int find(int x) { return fa[x] == x ? x : (fa[x] = find(fa[x])); }
int main() {
  N = read();
  ++cnt[0];
  for (int x, i = 1; i <= N; ++i) {
    x = read();
    ++cnt[x];
    sum -= x;
  }
  for (int i = 0; i < _; ++i) {
    fa[i] = i;
    sz[i] = 1;
  }
  for (int i = _ - 1; i; --i) {
    int s = i;
    do {
      if (cnt[s] && cnt[i ^ s]) {
        int p = find(s), q = find(i ^ s);
        if (p == q) continue;
        sum += 1ll * (cnt[p] + cnt[q] - 1) * i;
        if (sz[p] > sz[q]) swap(p, q);
        fa[p] = q;
        cnt[q] = 1;
        sz[q] += sz[p];
      }
    } while (s = (s - 1) & i);
  }
  cout << sum;
  return 0;
}