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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int maxn = 1 << 18;
int p[maxn], a[maxn];
int dp[maxn][2];
int root(int v) { return v == p[v] ? v : (p[v] = root(p[v])); }
template <class T>
void intro(T l, int i) {
  if (i == -1) return;
  int p = root(i);
  for (int z = 0; z < (int)(2); ++z) {
    if (l[z] == -1) {
      l[z] = i;
      break;
    }
    if (root(l[z]) == p) {
      if (a[i] > a[l[z]]) l[z] = i;
      break;
    } else {
      if (a[i] > a[l[z]]) swap(l[z], i), p = root(i);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  a[n] = 0;
  ++n;
  for (int i = 0; i < (int)(n); ++i) p[i] = i;
  long long ans = 0;
  for (int k = 0; k < (int)(20); ++k) {
    for (int j = 0; j < (int)(maxn); ++j)
      for (int z = 0; z < (int)(2); ++z) dp[j][z] = -1;
    for (int i = 0; i < (int)(n); ++i) intro(dp[a[i]], i);
    for (int j = 0; j < (int)(18); ++j)
      for (int i = 0; i < (int)(maxn); ++i)
        if (!((i >> j) & 1))
          for (int z = 0; z < (int)(2); ++z) intro(dp[i + (1 << j)], dp[i][z]);
    vector<pair<int, int> > best(n, {-1, -1});
    for (int i = 0; i < (int)(n); ++i) {
      int x = a[i] ^ (maxn - 1);
      for (int z = 0; z < (int)(2); ++z) {
        int y = dp[x][z];
        if (y != -1 && root(y) != root(i)) {
          int p = root(i);
          uax(best[p], {a[i] + a[y], y});
        }
      }
    }
    for (int i = 0; i < (int)(n); ++i) {
      int y = best[i].second;
      if (y == -1 || root(y) == root(i)) continue;
      ans += best[i].first;
      p[i] = root(y);
    }
  }
  for (int i = 0; i < (int)(n); ++i) ans -= a[i];
  cout << ans << '\n';
  return 0;
}