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(2)
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 1;
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  if (f) x = -x;
}
template <typename F>
inline void write(F x) {
  static short st[30];
  short tp = 0;
  if (x < 0) putchar('-'), x = -x;
  do st[++tp] = x % 10, x /= 10;
  while (x);
  while (tp) putchar('0' | st[tp--]);
  putchar('\n');
}
template <typename T>
inline void Mx(T &x, T y) {
  x < y && (x = y);
}
template <typename T>
inline void Mn(T &x, T y) {
  x > y && (x = y);
}
const int lim = 1 << 18, N = lim + 5;
long long sum, n;
int f[N], cnt[N];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int main() {
  read(n);
  for (int i = 1, x; i <= n; i++) read(x), cnt[x]++, sum -= x;
  for (int i = 1; i <= lim; i++) f[i] = i;
  cnt[0]++;
  for (int i = lim - 1; i >= 0; i--) {
    for (int j = i; j; j = (j - 1) & i) {
      if (!cnt[j] || !cnt[i ^ j]) continue;
      int x = find(j), y = find(i ^ j);
      if (x == y) continue;
      f[y] = x, sum += (long long)(cnt[j] + cnt[i ^ j] - 1) * i;
      cnt[j] = cnt[i ^ j] = 1;
    }
  }
  write(sum);
  return 0;
}