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;
namespace IO {
const int BUFFER_SIZE = 1 << 15;
char input_buffer[BUFFER_SIZE];
int input_pos = 0, input_len = 0;
char output_buffer[BUFFER_SIZE];
int output_pos = 0;
char number_buffer[100];
uint8_t lookup[100];
void _update_input_buffer() {
  input_len = fread(input_buffer, sizeof(char), BUFFER_SIZE, stdin);
  input_pos = 0;
  if (input_len == 0) input_buffer[0] = EOF;
}
inline char next_char(bool advance = true) {
  if (input_pos >= input_len) _update_input_buffer();
  return input_buffer[advance ? input_pos++ : input_pos];
}
template <typename T>
inline void read_int(T &number) {
  bool negative = false;
  number = 0;
  while (!isdigit(next_char(false)))
    if (next_char() == '-') negative = true;
  do {
    number = 10 * number + (next_char() - '0');
  } while (isdigit(next_char(false)));
  if (negative) number = -number;
}
template <typename T, typename... Args>
inline void read_int(T &number, Args &...args) {
  read_int(number);
  read_int(args...);
}
void _flush_output() {
  fwrite(output_buffer, sizeof(char), output_pos, stdout);
  output_pos = 0;
}
inline void write_char(char c) {
  if (output_pos == BUFFER_SIZE) _flush_output();
  output_buffer[output_pos++] = c;
}
template <typename T>
inline void write_int(T number, char after = '\0') {
  if (number < 0) {
    write_char('-');
    number = -number;
  }
  int length = 0;
  while (number >= 10) {
    uint8_t lookup_value = lookup[number % 100];
    number /= 100;
    number_buffer[length++] = (lookup_value & 15) + '0';
    number_buffer[length++] = (lookup_value >> 4) + '0';
  }
  if (number != 0 || length == 0) write_char(number + '0');
  for (int i = length - 1; i >= 0; i--) write_char(number_buffer[i]);
  if (after) write_char(after);
}
void init() {
  bool exit_success = atexit(_flush_output) == 0;
  assert(exit_success);
  for (int i = 0; i < 100; i++) lookup[i] = (i / 10 << 4) + i % 10;
}
}  // namespace IO
struct union_find {
  vector<int> p;
  union_find(int n) : p(n, -1) {}
  template <typename RAIter>
  int join(int u, int v, RAIter f) {
    if ((u = root(u)) == (v = root(v))) return 0;
    if (p[u] > p[v]) swap(u, v);
    int r = (size(u) == 1 ? f[u] : 1) + (size(v) == 1 ? f[v] : 1) - 1;
    p[u] += p[v];
    p[v] = u;
    return r;
  }
  int root(int u) { return p[u] < 0 ? u : p[u] = root(p[u]); }
  int size(int u) { return -p[root(u)]; }
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  IO::init();
  int n, sz = 18;
  IO::read_int(n);
  long long ans = 0;
  vector<int> a(n), f(1 << sz);
  for (auto &i : a) IO::read_int(i), ++f[i], ans -= i;
  ++f[0];
  union_find uf(1 << sz);
  for (int i = 1 << sz; --i >= 0;)
    for (int j = i; 2 * j >= i; (--j) &= i)
      if (f[i ^ j] && f[j]) {
        ans += (long long)i * uf.join(i ^ j, j, f.begin());
        if (j == 0) break;
      }
  IO::write_int(ans, '\n');
  return 0;
}