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 M = 18;
int fa[1 << M];
int cnt[1 << M];
int dep[1 << M];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int merge(int u, int v) {
  if (!cnt[u] || !cnt[v]) return 0;
  u = find(u);
  v = find(v);
  if (u == v) return 0;
  if (dep[u] > dep[v]) swap(u, v);
  int res = cnt[u] + cnt[v] - 1;
  fa[u] = v;
  cnt[v] = 1;
  if (dep[u] == dep[v]) dep[v]++;
  return res;
}
inline void re(int& x) {
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ '0');
    c = getchar();
  }
  return;
}
int main() {
  int n;
  re(n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    re(x);
    cnt[x]++;
    ans -= x;
  }
  cnt[0]++;
  for (int i = 0; i < (1 << M); i++) fa[i] = i, dep[i] = 0;
  for (int s = (1 << M) - 1; s; s--)
    for (int t = s; t; t = (t - 1) & s) {
      int r = s ^ t;
      ans += 1ll * s * merge(r, t);
    }
  printf("%lld\n", ans);
  return 0;
}