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 fa[400086], siz[400086];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int n;
int cnt[400086];
int x, y;
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    ++cnt[x];
    ans -= x;
  }
  for (int i = 0; i <= 2e5; i++) fa[i] = i, siz[i] = 1;
  ++cnt[0];
  for (int i = 1 << 18; i; i--) {
    for (int j = i;; j = (j - 1) & i) {
      int k = i ^ j;
      if (cnt[j] && cnt[k]) {
        x = find(j), y = find(k);
        if (x != y) {
          if (siz[x] > siz[y]) swap(x, y);
          siz[y] += siz[x];
          fa[x] = y;
          ans += 1ll * (cnt[j] + cnt[k] - 1) * i;
          cnt[j] = cnt[k] = 1;
        }
      }
      if (!j) break;
    }
  }
  printf("%lld", ans);
}