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 = 3e5 + 10, M = (1 << 18);
int n, a[N], fa[N], cnt;
long long ans;
pair<int, int> mx[M][2], cur[M];
int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]); }
bool merge(int x, int y) {
  x = getfa(x), y = getfa(y);
  if (x == y) return 0;
  fa[x] = y;
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]), fa[i] = i;
    ans -= a[i];
  }
  cnt = n + 1;
  while (cnt > 1) {
    for (int i = 0; i <= M - 1; i++)
      for (int j = 0; j <= 1; j++) mx[i][j] = make_pair(-1, -1);
    for (int i = 0; i <= n; i++) {
      pair<int, int> t = make_pair(a[i], getfa(i));
      if (mx[a[i]][0] < t) {
        if (mx[a[i]][0].second != t.second) mx[a[i]][1] = mx[a[i]][0];
        mx[a[i]][0] = t;
      } else if (mx[a[i]][1] < t) {
        if (mx[a[i]][0].second != t.second) mx[a[i]][1] = t;
      }
    }
    for (int i = 0; i <= M - 1; i++) {
      for (int j = 0; j <= 18; j++) {
        if (i & (1 << j)) {
          for (int k = 0; k <= 1; k++) {
            if (mx[i][0] < mx[i ^ (1 << j)][k]) {
              if (mx[i][0].second != mx[i ^ (1 << j)][k].second)
                mx[i][1] = mx[i][0];
              mx[i][0] = mx[i ^ (1 << j)][k];
            } else if (mx[i][1] < mx[i ^ (1 << j)][k]) {
              if (mx[i][0].second != mx[i ^ (1 << j)][k].second)
                mx[i][1] = mx[i ^ (1 << j)][k];
            }
          }
        }
      }
    }
    for (int i = 0; i <= n; i++) cur[i] = make_pair(-1, -1);
    for (int i = 0; i <= n; i++) {
      int f = getfa(i);
      int t = ((M - 1) ^ a[i]);
      if (mx[t][0].second != -1 && mx[t][0].second != f) {
        cur[f] = max(cur[f], make_pair(mx[t][0].first + a[i], mx[t][0].second));
      } else if (mx[t][1].second != -1 && mx[t][1].second != f) {
        cur[f] = max(cur[f], make_pair(mx[t][1].first + a[i], mx[t][1].second));
      }
    }
    for (int i = 0; i <= n; i++) {
      if (fa[i] == i && merge(i, cur[i].second)) {
        cnt--, ans += cur[i].first;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}