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 = 262144;
int n, x, c[N], f[N];
int F(int x) { return f[x] == x ? x : (f[x] = F(f[x])); }
long long ans;
void mer(int x, int y) {
  if (!c[x] || !c[y]) return;
  long long w = x + y;
  x = F(x), y = F(y);
  if (x != y) {
    f[x] = y;
    ans += (long long)w * (c[x] + c[y] - 1);
    c[y] = 1;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, _b = n; i <= _b; i++) scanf("%d", &x), c[x]++, ans -= x;
  c[0]++;
  for (int i = 0, _b = 2e5; i <= _b; i++) f[i] = i;
  for (int i = N - 1, _b = 1; i >= _b; i--) {
    for (int j = i; j; j = (j - 1) & i) {
      mer(j, i ^ j);
    }
  }
  printf("%lld\n", ans);
}