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 + 5;
int n;
int a[N], f[N], cnt[N], fa[N];
long long ans;
inline int findf(int x) { return x == fa[x] ? x : fa[x] = findf(fa[x]); }
inline void merge(int u, int v, int w) {
  u = findf(u), v = findf(v);
  if (u == v) return;
  ans += 1ll * (cnt[u] + cnt[v] - 1) * w;
  fa[u] = v, cnt[u] = cnt[v] = 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ans -= a[i], cnt[a[i]]++;
  for (int i = 0; i < (1 << 18); i++) fa[i] = i;
  cnt[0]++;
  for (int i = (1 << 18) - 1; i >= 0; i--) {
    for (int s = i; s; s = (s - 1) & i)
      if (cnt[s] && cnt[i ^ s]) merge(s, i ^ s, i);
  }
  printf("%lld", ans);
  return 0;
}