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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int maxn = (1 << 18);
int n, a[maxn], cnt[maxn], par[maxn], sizee[maxn];
bool mark[maxn];
long long ans = 0;
int get_par(int v) {
  if (par[v] == v)
    return v;
  else
    return par[v] = get_par(par[v]);
}
void dsu(int v, int u, int w) {
  v = get_par(v);
  u = get_par(u);
  if (v == u) return;
  if (sizee[v] < sizee[u]) swap(v, u);
  sizee[v] += sizee[u];
  par[u] = v;
  ans += w * 1ll *
         (1 + ((cnt[v] - 1) * 1ll * mark[v]) + ((cnt[u] - 1) * 1ll * mark[u]));
  mark[v] = mark[u] = false;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    par[a[i]] = a[i];
    sizee[a[i]] = 1;
    cnt[a[i]]++;
    mark[a[i]] = true;
    ans -= a[i];
  }
  cnt[0]++;
  for (int mask = maxn - 1; mask > 0; mask--) {
    for (int submask = mask; submask > 0; (--submask) &= mask) {
      int v = submask, u = mask ^ submask;
      if (cnt[v] > 0 && cnt[u] > 0) dsu(v, u, v + u);
    }
  }
  printf("%lld", ans);
  return 0;
}