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 maxn = 1 << 18;
int n;
int cnt[maxn];
bool vis[maxn];
namespace us {
int fa[maxn];
void init(int n) {
  for (int i = 0; i <= n; ++i) fa[i] = i;
}
int find(int a) { return a == fa[a] ? a : fa[a] = find(fa[a]); }
int merge(int a, int b) {
  a = find(a), b = find(b);
  if (a == b) return 0;
  int re = -1;
  re += vis[a] ? 1 : cnt[a];
  vis[a] = 1;
  re += vis[b] ? 1 : cnt[b];
  vis[b] = 1;
  fa[a] = b;
  return re;
}
}  // namespace us
int main() {
  scanf("%d", &n);
  long long an = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ++cnt[x];
    an -= x;
  }
  ++cnt[0];
  us::init((1 << 18) - 1);
  for (int s = (1 << 18) - 1; ~s; --s) {
    for (int t = s;; t = (t - 1) & s) {
      if (cnt[t] && cnt[s ^ t]) {
        an += (long long)s * us::merge(t, s ^ t);
      }
      if (t == 0) break;
    }
  }
  printf("%I64d\n", an);
  return 0;
}