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;
long long ans = 0;
int a[200005], n, len = 1 << 18;
int mx1[1 << 18], mx2[1 << 18], fa[200005], w1[200005], w2[200005];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void add(int x, int w) {
  if (find(w) != find(mx1[x]) && a[w] > a[mx1[x]])
    mx2[x] = mx1[x], mx1[x] = w;
  else if (find(w) != find(mx1[x]) && a[w] > a[mx2[x]])
    mx2[x] = w;
}
int query(int x) {
  int t = (len - 1) ^ a[x];
  if (mx1[t] && find(mx1[t]) != find(x)) return mx1[t];
  if (mx2[t] && find(mx2[t]) != find(x)) return mx2[t];
  return 0;
}
int merge(int x, int y) {
  if (y == 0) return 0;
  y = find(y);
  if (x == y) return 0;
  fa[x] = y;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), ans -= a[i];
  ++n;
  a[0] = -1;
  for (int i = 1; i <= n; ++i) fa[i] = i;
  while (1) {
    for (int i = 0; i < len; ++i) mx1[i] = mx2[i] = 0;
    for (int i = 1; i <= n; ++i) add(a[i], i);
    for (int i = 0; i < 18; ++i)
      for (int j = 0; j < len; ++j)
        if (j & (1 << i)) {
          add(j, mx1[j ^ (1 << i)]);
          add(j, mx2[j ^ (1 << i)]);
        }
    for (int i = 1; i <= n; ++i) w1[i] = -1;
    for (int i = 1; i <= n; ++i) {
      int t = query(i), x = find(i);
      if (t != 0 && a[i] + a[t] > w1[x]) w1[x] = a[i] + a[t], w2[x] = t;
    }
    int f = 0;
    for (int i = 1; i <= n; ++i)
      if (find(i) == i && merge(i, w2[i])) ans += w1[i], f = 1;
    if (!f) break;
  }
  printf("%lld", ans);
}