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>
int n, cnt[2000005], fa[2000005];
long long ans;
int read() {
  char c = getchar();
  int ans = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans;
}
void Write(long long x) {
  if (x < 10)
    putchar(x ^ 48);
  else
    Write(x / 10), putchar((x % 10) ^ 48);
  return;
}
int found(int x) { return fa[x] == x ? x : fa[x] = found(fa[x]); }
int main() {
  n = read();
  cnt[0] = 1;
  for (register int i = 1, x; i <= n; i++) ++cnt[x = read()], ans -= x;
  for (register int i = 0; i < (1 << 18); i++) fa[i] = i;
  for (register int i = (1 << 18) - 1; i > 0; i--)
    for (register int j = i; j; j = (j - 1) & i) {
      int k = i ^ j;
      if (!cnt[j] || !cnt[k] || found(j) == found(k)) continue;
      int x = found(j), y = found(k);
      fa[found(x)] = found(y);
      ans += (i + 0ll) * (cnt[x] + cnt[y] - 1), cnt[x] = cnt[y] = 1;
    }
  Write(ans);
  return 0;
}