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 void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
int n, m, i, j, a[200005], fa[262145], c[262145];
long long ans;
int fnd(int x) {
  if (fa[x] == x) return x;
  return fa[x] = fnd(fa[x]);
}
int main() {
  read(n);
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) read(a[i]);
  c[0] = 1;
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
    ans -= a[i];
    c[a[i]]++;
  }
  for (((i)) = (0); ((i)) <= (((int)((1 << 18))) - 1); ((i))++) fa[i] = i;
  for (i = (1 << 18) - 1; i; i--) {
    for (j = i; j; j = ((j - 1) & i)) {
      if (c[j] && c[i ^ j] && fnd(j) != fnd(i ^ j)) {
        int x = fnd(j), y = fnd(i ^ j);
        ans += 1ll * i * (c[x] + c[y] - 1);
        c[x] = c[y] = 1;
        fa[x] = y;
      }
    }
  }
  cout << ans << endl;
  return 0;
}