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;
const int maxn = 2e5 + 5;
const int mlog = 18;
const int all = (1 << 18);
const int inf = 1e9;
int n;
int a[maxn];
int head[maxn];
pair<int, int> dp[all + 5][2];
pair<int, int> to[maxn];
long long ans;
void update(int bit, pair<int, int> val) {
  if (dp[bit][0].second == val.second) {
    dp[bit][0] = max(dp[bit][0], val);
  } else if (dp[bit][1].second == val.second) {
    dp[bit][1] = max(dp[bit][1], val);
    if (dp[bit][1] > dp[bit][0]) swap(dp[bit][0], dp[bit][1]);
  } else {
    for (int i = 0; i < 2; i++)
      if (dp[bit][i] < val) swap(dp[bit][i], val);
  }
}
int findhead(int x) {
  if (head[x] == x) return x;
  return head[x] = findhead(head[x]);
}
int edge(int x, int y, int val) {
  x = findhead(x);
  y = findhead(y);
  if (x == y) return 0;
  head[x] = y;
  return val;
}
void mst() {
  for (int bit = 0; bit < all; bit++) dp[bit][0] = dp[bit][1] = {-inf, 0};
  for (int i = 1; i <= n; i++) update(a[i], {a[i], findhead(i)});
  for (int i = 0; i < mlog; i++)
    for (int mask = 0; mask < all; mask++) {
      if (mask & (1 << i)) {
        for (int j = 0; j < 2; j++) update(mask, dp[mask ^ (1 << i)][j]);
      }
    }
  for (int x = 1; x <= n; x++) to[x] = {-inf, 0};
  for (int x = 1; x <= n; x++) {
    int mask = all - 1 - a[x];
    pair<int, int> mx =
        (dp[mask][0].second != findhead(x) ? dp[mask][0] : dp[mask][1]);
    to[findhead(x)] = max(to[findhead(x)], {mx.first + a[x], mx.second});
  }
  for (int x = 1; x <= n; x++)
    if (x == findhead(x)) ans += edge(x, to[x].second, to[x].first);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[++n] = 0;
  for (int i = 1; i <= n; i++) head[i] = i;
  while (1) {
    int cnt = 0;
    for (int i = 1; i <= n; i++) cnt += i == findhead(i);
    if (cnt == 1) break;
    mst();
  }
  for (int i = 1; i <= n; i++) ans -= a[i];
  printf("%lld", ans);
}