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 = 270100;
const int M = (1 << 18) - 1;
inline int read() {
  char ch;
  while ((ch = getchar()) < '0' || ch > '9')
    ;
  int res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return res;
}
int n, f[N], cnt[N];
inline int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); }
int main() {
  n = read();
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x = read();
    cnt[x]++, ans -= x;
  }
  cnt[0]++;
  for (int i = 0; i <= M; i++) f[i] = i;
  for (int i = M; i >= 1; i--) {
    int u, v;
    for (int j = i;; j = (j - 1) & i) {
      if (cnt[j] && cnt[i ^ j]) {
        int u = find(j), v = find(i ^ j);
        if (u != v) {
          f[u] = v;
          ans += 1ll * (cnt[j] + cnt[i ^ j] - 1) * i;
          cnt[j] = 1;
          cnt[i ^ j] = 1;
        }
      }
      if (j == 0) break;
    }
  }
  printf("%lld", ans);
  return 0;
}