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 n, ma = 2e5, af = 1 << 18, fq[1 << 18], dsu[200069];
int fd(int x) {
  if (dsu[x] != x) {
    dsu[x] = fd(dsu[x]);
  }
  return dsu[x];
}
int main() {
  int i, k, l, mk, mkk;
  long long z = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &k);
    fq[k]++;
    z -= k;
  }
  fq[0]++;
  for (i = 0; i <= ma; i++) {
    dsu[i] = i;
  }
  for (mk = af - 1; mk; mk--) {
    for (mkk = mk; mkk > (mk ^ mkk); mkk = mkk - 1 & mk) {
      if (fq[mkk] && fq[mk ^ mkk]) {
        k = fd(mkk);
        l = fd(mk ^ mkk);
        if (k != l) {
          z += (long long)mk * (fq[mkk] + fq[mk ^ mkk] - 1);
          dsu[l] = k;
          fq[k] = 1;
          fq[l] = 1;
        }
      }
    }
  }
  printf("%lld\n", z);
}