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;
inline int rd() {
  int x = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' && c != '-') c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0') x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return f ? -x : x;
}
const int SIZE = 1 << 18;
int n;
int fa[SIZE], a[SIZE];
int Find(int x) {
  if (x == fa[x]) return x;
  return fa[x] = Find(fa[x]);
}
void _main() {
  n = rd();
  long long ans = 0;
  ++a[0];
  for (register int i = 1, _n = n; i <= _n; ++i) {
    int x = rd();
    ++a[x];
    ans -= x;
  }
  for (register int i = 0, _n = (1 << 18); i < _n; ++i) fa[i] = i;
  for (register int i = (1 << 18) - 1, _n = 1; i >= _n; --i) {
    int S = i & (i - 1), T = i & -i;
    for (int j = S;; j = (j - 1) & S) {
      if (a[j] && a[i ^ j]) {
        int x = Find(j), y = Find(i ^ j);
        if (x != y) {
          fa[x] = y;
          ans += 1LL * (a[x] + a[y] - 1) * i;
          a[y] = 1;
        }
      }
      if (j == 0) break;
    }
  }
  printf("%lld\n", ans);
}
int main() {
  _main();
  return 0;
}