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 = 2e5 + 10;
const int mx = (1 << 18);
int n;
int cnt[mx], p[mx], rnk[mx];
long long ans;
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
inline int get_root(int v) { return (p[v] == v) ? v : (p[v] = get_root(p[v])); }
inline int join(int u, int v) {
  u = get_root(u), v = get_root(v);
  if (u != v) {
    if (rnk[u] > rnk[v]) swap(u, v);
    p[u] = v;
    if (rnk[u] == rnk[v]) ++rnk[v];
    return 1;
  } else
    return 0;
}
int main() {
  read(n);
  for (int i = 0; i < n; ++i) {
    int val;
    read(val);
    ans -= val;
    ++cnt[val];
  }
  iota(p, p + mx, 0);
  ++cnt[0];
  for (int e = mx - 1; e >= 0; --e)
    for (int i = e; true; i = ((i - 1) & e)) {
      if (cnt[i] && cnt[e ^ i] && get_root(i) != get_root(e ^ i)) {
        join(i, e ^ i);
        ans += (long long)(cnt[i] + cnt[e ^ i] - 1) * e;
        cnt[i] = 1;
        cnt[i ^ e] = 1;
      }
      if (i == 0) break;
    }
  printf("%lld\n", ans);
  return 0;
}