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 has[300005];
int val[300005];
int Max[300005];
int f[300005];
int cnt[300005];
long long ans;
int Find(int first) {
  if (f[first] == first) return first;
  return f[first] = Find(f[first]);
}
int main() {
  int n;
  scanf("%d", &n);
  int a[200005];
  long long tmp = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    f[i] = i;
    has[a[i]] = 1;
    val[a[i]] = i;
    cnt[a[i]]++;
    ans += -a[i];
    tmp += -a[i];
  }
  a[n] = 0;
  f[n] = n;
  has[0] = 1;
  val[0] = n;
  cnt[0]++;
  for (int i = (1 << 18) - 1; i >= 0; i--) {
    for (int a = i; a; a = (a - 1) & i) {
      int b = i - a;
      if (has[a] && has[b]) {
        if (Find(val[a]) != Find(val[b])) {
          ans += (cnt[a] + cnt[b] - 1) * (long long)(i);
          cnt[a] = 1, cnt[b] = 1;
          f[Find(val[a])] = Find(val[b]);
        }
      }
    }
  }
  printf("%lld\n", ans);
}