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 N = (1 << 18) + 5;
int n;
long long res = 0;
int f[N], sz[N], cnt[N];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
long long unite(int x, int y) {
  x = find(x), y = find(y);
  if (x == y) return 0;
  long long res = (sz[x] == 1 ? cnt[x] : 1) + (sz[y] == 1 ? cnt[y] : 1) - 1;
  sz[x] += sz[y];
  f[y] = x;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cnt[x]++;
    res -= x;
  }
  cnt[0]++;
  for (int i = 0; i < (1 << 18); i++) f[i] = i, sz[i] = 1;
  for (int i = (1 << 18) - 1; i >= 0; i--) {
    if (cnt[i] > 0 && cnt[0] > 0) res += unite(i, 0) * i;
    for (int x = i; x;) {
      --x &= i;
      if (cnt[x] > 0 && cnt[i ^ x] > 0) {
        res += unite(x, i ^ x) * i;
      }
      if (x * 2 < i) break;
    }
  }
  cout << res << endl;
}