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;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 18;
int n, cnt;
int a[200005], f[200005];
pair<int, int> d[300005][2], now[200005];
int getf(int x) { return f[x] == x ? x : f[x] = getf(f[x]); }
bool merge(int x, int y) {
  int fx = getf(x), fy = getf(y);
  if (fx == fy) return false;
  f[fx] = fy, cnt++;
  return true;
}
void upd(int x, pair<int, int> t) {
  if (t.second == d[x][0].second)
    chkmax(d[x][0].first, t.first);
  else if (!chkmax(d[x][0], t))
    chkmax(d[x][1], t);
}
int main() {
  n = readint();
  long long ans = 0;
  for (int i = 1; i <= n; i++) a[i] = readint(), f[i] = i, ans -= a[i];
  while (1) {
    for (int i = 0; i < (1 << N); i++) d[i][0] = d[i][1] = make_pair(0, 0);
    for (int i = 1; i <= n; i++) upd(a[i], make_pair(a[i], getf(i)));
    for (int i = 0; i < N; i++)
      for (int j = 0; j < (1 << N); j++)
        if (j & (1 << i))
          upd(j, d[j ^ (1 << i)][0]), upd(j, d[j ^ (1 << i)][1]);
    for (int i = 1; i <= n; i++) now[i] = make_pair(0, 0);
    for (int i = 0; i <= n; i++) {
      int tmp = a[i] ^ ((1 << N) - 1);
      if (d[tmp][0].second == getf(i))
        chkmax(now[f[i]], make_pair(d[tmp][1].first + a[i], d[tmp][1].second));
      else
        chkmax(now[f[i]], make_pair(d[tmp][0].first + a[i], d[tmp][0].second));
    }
    for (int i = 0; i <= n; i++)
      if (f[i] == i && merge(i, now[i].second)) ans += now[i].first;
    if (cnt == n) break;
  }
  printf("%lld\n", ans);
  return 0;
}