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 a[500010];
int pre[500010];
long long val[500010];
int find(int x) { return pre[x] == x ? x : pre[x] = find(pre[x]); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int tmp;
    cin >> tmp;
    a[tmp]++;
    ans -= tmp;
  }
  a[0]++;
  for (int i = 0; i <= 200000; i++) {
    pre[i] = i;
  }
  const int N = 1 << 18;
  for (int sum = N - 1; sum >= 0; sum--) {
    for (int u = sum; u > (sum ^ u); (--u) &= sum) {
      int v = sum ^ u;
      if (a[v] && a[u]) {
        int fi = find(v);
        int fj = find(u);
        if (fi != fj) {
          val[fj] += val[fi] + sum;
          pre[fi] = fj;
        }
      }
    }
  }
  for (int i = 200000; i >= 0; i--) {
    if (!a[i]) continue;
    int k = ((1 << 18) - 1) ^ i;
    int flg = 0;
    for (int j = k;; j = (j - 1) & k) {
      if (a[j]) {
        ans += 1ll * (a[i] - 1) * (i + j);
        break;
      }
      if (!j) break;
    }
  }
  cout << ans + val[find(0)] << endl;
}