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;
int fa[(1 << 18) + 5], cnt[(1 << 18) + 5], h[(1 << 18) + 5];
long long ans;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int x, int y) {
  if (x == y || !cnt[x] || !cnt[y]) return;
  int t = (x | y);
  x = find(x), y = find(y);
  if (x == y) return;
  if (h[x] > h[y]) swap(x, y);
  fa[x] = y, h[y] = max(h[y], h[x] + 1), ans += 1ll * t * (cnt[x] + cnt[y] - 1),
  cnt[y] = 1;
}
int main() {
  int n, x;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &x), ans -= x, cnt[x]++;
  cnt[0]++;
  for (int i = 0; i < (1 << 18); i++) fa[i] = i, h[i] = 1;
  for (int s = (1 << 18) - 1; s; s--)
    for (int t = s; t > (s ^ t); t = ((t - 1) & s)) merge(t, s ^ t);
  printf("%lld", ans);
  return 0;
}