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>
inline void read(T &x) {
  int ch = 0, f = 0;
  x = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  if (f) x = -x;
}
const int N = 1 << 18;
long long ans;
pair<int, int> dp[N][2], chs[N];
int a[N], fa[N], n;
inline int ask(int x) { return x == fa[x] ? x : fa[x] = ask(fa[x]); }
inline void update(int x, pair<int, int> y) {
  if (dp[x][0] < y) {
    if (ask(y.second) != ask(dp[x][0].second)) dp[x][1] = dp[x][0];
    dp[x][0] = y;
  } else if (dp[x][1] < y && ask(y.second) != ask(dp[x][0].second))
    dp[x][1] = y;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]), fa[i] = i;
  ++n, fa[n] = n;
  int block = n;
  while (block > 1) {
    for (int i = 0; i < N; i++) {
      dp[i][0].first = -1, dp[i][0].second = 0;
      dp[i][1].first = -1, dp[i][1].second = 0;
    }
    for (int i = 1; i <= n; i++) chs[i].first = chs[i].second = 0;
    for (int i = 1; i <= n; i++) update(a[i], make_pair(a[i], i));
    for (int i = 0; i < 18; i++)
      for (int j = 0; j < N; j++)
        if ((1 << i) & j) {
          update(j, dp[j ^ (1 << i)][0]);
          update(j, dp[j ^ (1 << i)][1]);
        }
    for (int i = 1; i <= n; i++) {
      int x = dp[(N - 1) ^ a[i]][0].second;
      if (!x) continue;
      if (ask(x) == ask(i)) x = dp[(N - 1) ^ a[i]][1].second;
      if (!x) continue;
      int xx = chs[ask(i)].first;
      int yy = chs[ask(i)].second;
      if (a[i] + a[x] > a[xx] + a[yy] || !xx) chs[ask(i)] = make_pair(i, x);
    }
    for (int i = 1; i <= n; i++)
      if (chs[i].first) {
        int p = ask(chs[i].first), q = ask(chs[i].second);
        if (p == q) continue;
        block--;
        ans += a[chs[i].first] + a[chs[i].second];
        fa[p] = q;
      }
  }
  for (int i = 1; i <= n; i++) ans -= a[i];
  cout << ans << endl;
  return 0;
}