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;
int n, cnt[N], dsu[N];
long long ans;
int trace(int x) { return dsu[x] < 0 ? x : dsu[x] = trace(dsu[x]); }
int merge(int x, int y) {
  x = trace(x), y = trace(y);
  if (x == y) return 0;
  if (dsu[x] > dsu[y]) swap(x, y);
  int ret = (dsu[x] == -1 ? cnt[x] : 1) + (dsu[y] == -1 ? cnt[y] : 1) - 1;
  dsu[x] += dsu[y];
  dsu[y] = x;
  return ret;
}
int main() {
  scanf("%d", &n);
  cnt[0]++;
  fill(dsu, dsu + N, -1);
  for (int i = 1, a; i <= n; i++) scanf("%d", &a), cnt[a]++, ans -= a;
  for (int i = N - 1; i >= 0; i--)
    for (int j = i; j > 0; (--j) &= i)
      if (cnt[i ^ j] > 0 && cnt[j] > 0) ans += 1ll * i * merge(i ^ j, j);
  printf("%lld\n", ans);
  return 0;
}