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;
namespace jyy {
const int mod = 998244353, U = 262144;
const double eps = 1e-8;
inline int read() {
  static int x = 0, f = 1;
  x = 0, f = 1;
  static char ch = getchar();
  while (ch < '0' || ch > '9') {
    (ch == '-') ? f *= -1 : f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + int(ch - 48), ch = getchar();
  return x * f;
}
}  // namespace jyy
using namespace jyy;
long long n, ans, c[U], fa[U];
int find(int x) { return fa[x] != x ? fa[x] = find(fa[x]) : x; }
int merge(int x, int y) {
  int fx = find(x), fy = find(y);
  if (fx == fy) return 0;
  int res = c[x] + c[y] - 1;
  fa[fx] = fy, c[x] = c[y] = 1;
  return res;
}
int main() {
  n = read();
  for (int i = 1, x; i <= n; i++) x = read(), c[x]++, ans -= x;
  c[0]++;
  for (int i = 0; i < U; i++) fa[i] = i;
  for (int i = U - 1; i > 0; i--)
    for (int t = i; t > i / 2; t = (t - 1) & i) {
      if (c[t] && c[t ^ i]) ans += 1ll * merge(t, t ^ i) * i;
    }
  cout << ans;
}