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;
int n, cnt[N], par[N];
long long ans;
int get_par(int u) { return par[u] < 0 ? u : par[u] = get_par(par[u]); }
bool merge(int u, int v) {
  if ((u = get_par(u)) == (v = get_par(v))) return false;
  if (par[u] < par[v]) swap(u, v);
  return par[u] += par[v], par[v] = u, true;
}
int main() {
  cnt[0]++;
  scanf("%d", &n);
  memset(par, -1, sizeof par);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++, ans -= x;
  }
  for (int i = N - 1; i; i--)
    for (int j = i; j; --j &= i)
      if (cnt[j] && cnt[i ^ j] && merge(j, i ^ j))
        ans += (cnt[j] + cnt[i ^ j] - 1LL) * i, cnt[j] = cnt[i ^ j] = 1;
  printf("%lld", ans);
}