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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
const int lg = 18;
const int N = 1 << lg | 5;
int sub[N], par[N], cnt[N], n;
int find(int u) { return par[u] == u ? u : par[u] = find(par[u]); }
inline int single(int u) { return par[u] == u and sub[u] == 1; }
int unite(int u, int v) {
  int ret = 0;
  if (single(u) and single(v)) {
    ret = cnt[u] + cnt[v] - 1;
  } else if (single(u)) {
    ret = cnt[u];
  } else if (single(v)) {
    ret = cnt[v];
  } else {
    ret = find(u) != find(v);
  }
  u = find(u), v = find(v);
  if (sub[u] > sub[v]) swap(u, v);
  if (u != v) {
    par[u] = v, sub[v] += sub[u];
  }
  return ret;
}
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  ll total = 0;
  scanf("%d", &n);
  for (int i = 0, x; i < n; i++) {
    scanf("%d", &x);
    cnt[x]++, total -= x;
  }
  cnt[0]++;
  iota(par, par + N, 0);
  fill(sub, sub + N, 1);
  for (int both = (1 << lg) - 1; both >= 0; both--) {
    for (int u = both; u > 0; u = (u - 1) & both) {
      int v = both ^ u;
      if (cnt[u] and cnt[v]) {
        total += (ll)both * unite(u, v);
      }
    }
  }
  printf("%lld\n", total);
  return 0;
}