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;
const long long mod = 998244353;
int cnt[N], fa[N];
int dsu(int x) {
  if (fa[x] == x) return x;
  return fa[x] = dsu(fa[x]);
}
long long ans;
void solve(int u, int v) {
  if (!cnt[u] || !cnt[v]) return;
  if (dsu(u) == dsu(v)) return;
  fa[dsu(u)] = dsu(v);
  ans += 1ll * (u | v) * (cnt[u] + cnt[v] - 1);
  cnt[u] = 1;
  cnt[v] = 1;
}
int main() {
  int n, x;
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &x);
    ans -= x;
    cnt[x]++;
  }
  cnt[0]++;
  for (int i = (0); i < (N); ++i) fa[i] = i;
  for (int i = (N)-1; i >= (0); --i) {
    for (int j = i; j > (i ^ j); j = (j - 1) & i) {
      solve(j, j ^ i);
    }
  }
  printf("%lld", ans);
}