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 = (1 << 18);
template <class Ruby_Rose>
inline void read(Ruby_Rose &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
}
int n, buc[N + 5], fa[N + 5];
long long ans;
inline int getfa(int x) { return (fa[x] == x) ? x : fa[x] = getfa(fa[x]); }
inline void merge(int i, int j, int w) {
  if (buc[i] && buc[j]) {
    int fu = getfa(i), fv = getfa(j);
    if (fu == fv) return;
    ans += 1ll * (buc[fu] + buc[fv] - 1) * w;
    fa[fu] = fv, buc[fv] = 1;
  }
}
int main() {
  read(n);
  for (register int i = 1, qa; i <= n; i++) read(qa), buc[qa]++, ans -= qa;
  for (register int i = 0; i <= N; i++) fa[i] = i;
  buc[N] = 1;
  for (register int i = (1 << 18) - 1; ~i; i--) {
    for (register int j = i;; j = ((j - 1) & i)) {
      merge(i ^ j, j, i);
      if (!j) break;
    }
    merge(i, N, i);
  }
  cout << ans << endl;
  return 0;
}