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;
enum { N = 200007, M = 1 << 18, Z };
int n, x, C[Z], S[Z];
int fi(int u) { return u == C[u] ? u : C[u] = fi(C[u]); }
int main() {
  scanf("%d", &n);
  long long a = 0;
  for (int i = *S = 1; i <= n; ++i) scanf("%d", &x), ++S[x], a -= x;
  for (int i = 1; i <= M; ++i) C[i] = i;
  for (int i = M; i >= 1; --i)
    for (int w = i & (i - 1), j = w; j >= 0; --j &= w) {
      if (S[j] && S[i ^ j]) {
        int x = fi(j), y = fi(i ^ j);
        if (x != y && S[x] && S[y])
          a += 1LL * i * (S[x] + S[y] - 1), C[x] = y, S[y] = 1;
      }
      if (!j) break;
    }
  printf("%lld", a);
  return 0;
}