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;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
const int S = 262144;
int cnt[S];
int a[200005];
int fa[S];
long long ans = 0;
int n;
int gfa(int x) {
  if (fa[x] == x) return x;
  return fa[x] = gfa(fa[x]);
}
void lk(int u, int v) {
  if (!cnt[u] || !cnt[v]) return;
  if (gfa(u) == gfa(v)) return;
  int us = gfa(u), vs = gfa(v);
  fa[us] = vs;
  ans += 1ll * (u + v) * (cnt[u] + cnt[v] - 1);
  cnt[u] = 1, cnt[v] = 1;
}
int main() {
  int test = 0;
  if (test)
    n = 200000;
  else
    cin >> n;
  for (int i = 1; i <= n; i++)
    if (test)
      a[i] = i;
    else
      scanf("%d", &a[i]);
  a[++n] = 0;
  sort(a + 1, a + n + 1);
  for (int i = n; i >= 1; i--) cnt[a[i]]++;
  for (int i = 0; i < S; i++) fa[i] = i;
  for (int i = S - 1; i >= 0; i--) {
    for (int j = i; 1; j = (j - 1) & i) {
      lk(j, i ^ j);
      if (j == 0) break;
      if ((i ^ j) > j) break;
    }
  }
  for (int i = 1; i <= n; i++) ans -= a[i];
  cout << ans << endl;
  if (test) cout << clock() << endl;
  return 0;
}