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;
int a[(1 << 18) + 5], f[(1 << 18) + 5], fa[(1 << 18) + 5];
long long sf[(1 << 18) + 5];
int find(int x) { return x == fa[x] ? fa[x] : (fa[x] = find(fa[x])); }
int g[(1 << 18) + 5][2], h[(1 << 18) + 5][2];
int main() {
  int n;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n + 1; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ans -= a[i];
  }
  int cnt = n;
  while (cnt) {
    for (int i = 0; i < (1 << 18); i++)
      g[i][0] = g[i][1] = h[i][0] = h[i][1] = -1;
    for (int i = 1; i <= n + 1; i++) {
      int u = find(i);
      if (h[a[i]][0] == -1)
        g[a[i]][0] = a[i], h[a[i]][0] = u;
      else if (h[a[i]][1] == -1 && h[a[i]][0] != u)
        g[a[i]][1] = a[i], h[a[i]][1] = u;
    }
    for (int i = 0; i < 18; i++)
      for (int s = 0; s < (1 << 18); s++)
        if ((s >> i) & 1) {
          int t = s ^ (1 << i);
          for (int _ = 0; _ <= 1; _++)
            if (g[t][_] > g[s][0] ||
                (g[t][_] == g[s][0] && h[t][_] != h[s][0])) {
              if (h[t][_] != h[s][0]) g[s][1] = g[s][0], h[s][1] = h[s][0];
              g[s][0] = g[t][_], h[s][0] = h[t][_];
            } else if (g[t][_] > g[s][1] && h[t][_] != h[s][0])
              g[s][1] = g[t][_], h[s][1] = h[t][_];
        }
    for (int i = 1; i <= n + 1; i++) sf[i] = f[i] = -1;
    for (int i = 1; i <= n + 1; i++) {
      int s = ((1 << 18) - 1) ^ a[i], t = find(i);
      if (g[s][0] == -1) continue;
      for (int _ = 0; _ <= 1; _++)
        if ((g[s][_] ^ a[i]) > sf[t] && h[s][_] != t)
          f[t] = h[s][_], sf[t] = g[s][_] ^ a[i];
    }
    for (int i = 1; i <= n + 1; i++)
      if (i == find(i) && sf[i] >= 0 && find(f[i]) != find(i)) {
        fa[find(i)] = find(f[i]);
        --cnt;
        ans += sf[i];
      }
  }
  printf("%lld\n", ans);
  return 0;
}