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>
const int N = 1 << 18 | 1;
int fa[N], sz[N], rk[N];
inline int ancestor(int x) { return x == fa[x] ? x : fa[x] = ancestor(fa[x]); }
inline bool merge(int x, int y) {
  x = ancestor(x), y = ancestor(y);
  if (x == y) return 0;
  if (rk[x] < rk[y]) std::swap(x, y);
  fa[y] = x;
  if (rk[x] == rk[y]) ++rk[x];
  return 1;
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0), std::cout.tie(0);
  int n;
  long long ret = 0;
  std::cin >> n;
  for (int i = 0, iend = N - 1; i <= iend; ++i) fa[i] = i;
  sz[0] = 1;
  for (int i = 1, iend = n; i <= iend; ++i) {
    int x;
    std::cin >> x;
    ++sz[x], ret -= x;
  }
  for (int i = (1 << 18) - 1, iend = 0; i >= iend; --i) {
    for (int j = i; j; j = j - 1 & i) {
      if (sz[j] && sz[j ^ i] && merge(j, j ^ i)) {
        ret += 1ll * (sz[j] + sz[j ^ i] - 1) * i;
        sz[j] = sz[j ^ i] = 1;
      }
    }
  }
  std::cout << ret << '\n';
  return 0;
}