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>
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  register bool f = 0;
  while (!isdigit(c)) f ^= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
const int N = 2.7e5;
int n, max, lim, a[N], fa[N], cnt[N];
long long ans;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int u, int v, long long w) {
  if (cnt[u] && cnt[v]) {
    u = find(u), v = find(v);
    if (u != v) {
      ans += w * (cnt[u] + cnt[v] - 1);
      fa[u] = v, cnt[v] = 1;
    }
  }
}
int main() {
  read(n), cnt[0] = 1;
  for (int i = 1; i <= n; i++) {
    read(a[i]), cnt[a[i]]++;
    ans -= a[i];
    max = std::max(max, a[i]);
  }
  int lim = 1 << 32 - __builtin_clz(max);
  for (int i = 0; i < lim; i++) fa[i] = i;
  for (int i = lim - 1; ~i; i--) {
    for (int j = i; j; j = (j - 1) & i)
      if (cnt[j] && cnt[i ^ j]) merge(j, i ^ j, i);
    merge(i, 0, i);
  }
  printf("%lld\n", ans);
}