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 ans = 0;
int num[200010], from[300010], size[300010], n;
inline int rd() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
inline int find(int x) { return (x == from[x]) ? x : from[x] = find(from[x]); }
inline void merge(int x, int y, int z) {
  if (!size[x] || !size[y]) return;
  x = find(x);
  y = find(y);
  if (x != y) {
    ans += (long long)z * (size[x] + size[y] - 1);
    from[x] = y;
    size[y] = 1;
  }
}
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) num[i] = rd();
  for (int i = 0; i <= 262144; i++) from[i] = i;
  for (int i = 1; i <= n; i++) size[num[i]]++, ans -= num[i];
  size[262144] = 1;
  for (int i = 262143; ~i; i--) {
    for (int j = i; j; j = (j - 1) & i) merge(j, i ^ j, i);
    merge(i, 262144, i);
  }
  printf("%lld\n", ans);
  return 0;
}