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) + 100;
int n, a[N], lim, fa[N], g[N], h[N], size[N];
pair<int, int> f[N];
priority_queue<pair<int, int> > que;
template <class T>
inline void read(T &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
  return;
}
int find(int x) {
  if (x < 0) return -1;
  return x == fa[x] ? x : fa[x] = find(fa[x]);
}
inline pair<int, int> merge(pair<int, int> a, pair<int, int> b) {
  pair<int, int> ret(-1, -1);
  ret.first = max(a.first, b.first);
  if (find(ret.first) != find(a.first) && a.first > ret.second)
    ret.second = a.first;
  if (find(ret.first) != find(a.second) && a.second > ret.second)
    ret.second = a.second;
  if (find(ret.first) != find(b.first) && b.first > ret.second)
    ret.second = b.first;
  if (find(ret.first) != find(b.second) && b.second > ret.second)
    ret.second = b.second;
  return ret;
}
int main() {
  int x;
  read(n);
  long long ans = 0;
  for (register int i = 1; i <= n; ++i) {
    read(x), ++a[x], fa[x] = x;
    while (1 << lim <= x) ++lim;
    ans -= x, size[x] = 1;
  }
  ++a[0], size[0] = 1;
  for (register int i = 1; i <= lim; ++i) {
    for (register int j = 0; j < 1 << lim; ++j)
      f[j] = make_pair(a[j] ? j : -1, -1), g[j] = -1, h[j] = 0;
    for (register int j = 0; j < lim; ++j)
      for (register int k = 0; k < 1 << lim; ++k)
        if (k & (1 << j)) f[k] = merge(f[k], f[k ^ (1 << j)]);
    for (register int j = 0; j < 1 << lim; ++j) {
      if (!a[j]) continue;
      pair<int, int> w = f[j ^ ((1 << lim) - 1)];
      x = find(w.first) == find(j) ? w.second : w.first;
      if (h[find(j)] < j + x) h[find(j)] = j + x, g[find(j)] = x;
    }
    for (register int j = 0; j < 1 << lim; ++j)
      if (g[j] != -1) que.emplace(h[j], j);
    while (que.size()) {
      int j = que.top().second;
      que.pop();
      if (find(j) != find(g[j])) {
        long long x = size[j] == 1 ? a[j] : 1,
                  y = size[find(g[j])] == 1 ? a[g[j]] : 1;
        ans += (x + y - 1) * h[j], fa[find(g[j])] = j,
                                   size[j] += size[find(g[j])];
      }
    }
  }
  cout << ans << endl;
  return 0;
}