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;
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool d = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') d = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return d ? x : -x;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void writep(long long x) {
  write(x);
  putchar(' ');
}
inline unsigned long long rnd() {
  return ((unsigned long long)rand() << 30 ^ rand()) << 4 | rand() % 4;
}
const int N = (1 << 18) + 5;
int fa[N], size[N], cnt[N];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  int n = read();
  long long ans = 0;
  for (int i = (int)(1); i <= (int)(n); i++) {
    int x = read();
    cnt[x]++;
    ans -= x;
  }
  for (int i = (int)(0); i <= (int)(N - 1); i++) fa[i] = i, size[i] = 1;
  cnt[0]++;
  for (int s = (int)(1 << 18); s >= (int)(1); s--) {
    for (int i = s;; i = (i - 1) & s) {
      int j = s ^ i;
      if (cnt[i] && cnt[j]) {
        int u = find(i), v = find(j);
        if (u != v) {
          if (size[u] > size[v]) swap(u, v);
          size[v] += size[u];
          fa[u] = v;
          ans += (long long)(cnt[i] + cnt[j] - 1) * s;
          cnt[i] = cnt[j] = 1;
        }
      }
      if (i == 0) break;
    }
  }
  cout << ans;
}