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>
#pragma GCC optimize("Ofast")
using namespace std;
inline int read() {
  int ret = 0, t = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') t = -1, c = getchar();
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * t;
}
const int MAXN = 262145;
int n, a[MAXN], fa[MAXN];
int v[MAXN];
int gf(int x) {
  if (fa[x] == x) return x;
  return fa[x] = gf(fa[x]);
}
int main() {
  n = read();
  int i, j;
  ++v[0];
  long long ans = 0;
  for (i = 1; i <= n; i++) a[i] = read(), ++v[a[i]], ans -= a[i];
  for (i = 0; i <= 262144; i++) fa[i] = i;
  for (i = 262144; i >= 1; i--) {
    for (j = i; 1; j = i & (j - 1)) {
      int x = j, y = i ^ j;
      if (v[x] && v[y] && gf(x) != gf(y))
        fa[gf(x)] = gf(y), ans += 1ll * i * (v[x] + v[y] - 1), v[x] = v[y] = 1;
      if (!j) break;
    }
  }
  printf("%lld\n", ans);
  return 0;
}