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>
inline int read() {
  char c;
  int x;
  for (c = getchar(); !isdigit(c); c = getchar())
    ;
  for (x = 0; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x;
}
const int N = 1 << 20;
int n, lim, cnt[N];
long long ans;
struct DisjointSet {
  int anc[N], siz[N];
  void reset(int n) {
    for (int u = 0; u <= n; u++) {
      anc[u] = u;
      siz[u] = 1;
    }
  }
  int find(int u) { return anc[u] == u ? u : anc[u] = find(anc[u]); }
  bool merge(int u, int v) {
    u = find(u);
    v = find(v);
    if (u == v) {
      return false;
    }
    if (siz[u] < siz[v]) {
      anc[u] = v;
      siz[v] += siz[u];
    } else {
      anc[v] = u;
      siz[u] += siz[v];
    }
    return true;
  }
} ds;
int main() {
  n = read();
  lim = (1 << 18) - 1;
  cnt[0]++;
  ds.reset(lim);
  for (int i = 0; i < n; i++) {
    int x = read();
    cnt[x]++;
    ans -= x;
  }
  for (int i = lim; i > 0; i--) {
    for (int s = i, t;; s = (s - 1) & i) {
      t = i ^ s;
      if (cnt[s] > 0 && cnt[t] > 0 && ds.merge(s, t)) {
        ans += (cnt[s] + cnt[t] - 1ll) * i;
        cnt[s] = 1;
        cnt[t] = 1;
      }
      if (s == 0) {
        break;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}