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 N = (1 << 18);
int n;
int p[N];
long long a[N];
long long ans;
array<long long, 2> dp[N][2];
array<long long, 2> best[N];
int get(int x) { return (x == p[x]) ? x : p[x] = get(p[x]); }
bool unite(int x, int y) {
  x = get(x);
  y = get(y);
  if (x == y) {
    return 0;
  }
  p[y] = x;
  return 1;
}
void build_dp() {
  for (int i = 0; i < (1 << 18); ++i) {
    for (int j = 0; j < 2; ++j) {
      dp[i][j] = {-1, -1};
    }
    best[i] = {-1, -1};
  }
  for (int i = 0; i <= n; ++i) {
    array<long long, 2> cur = {a[i], get(i)};
    if (dp[a[i]][0][0] < cur[0]) {
      if (dp[a[i]][0][1] != cur[1]) {
        dp[a[i]][1] = dp[a[i]][0];
      }
      dp[a[i]][0] = cur;
    } else if ((dp[a[i]][1][0] < cur[0]) && (dp[a[i]][0][1] != cur[1])) {
      dp[a[i]][1] = cur;
    }
  }
  for (int bt = 0; bt < 18; ++bt) {
    for (int msk = 0; msk < (1 << 18); ++msk) {
      if (msk & (1 << bt)) {
        for (int it = 0; it < 2; ++it) {
          int nw = (msk ^ (1 << bt));
          if (dp[msk][0][0] < dp[nw][it][0]) {
            if (dp[msk][0][1] != dp[nw][it][1]) {
              dp[msk][1] = dp[msk][0];
            }
            dp[msk][0] = dp[nw][it];
          } else if ((dp[msk][1][0] < dp[nw][it][0]) &&
                     (dp[msk][0][1] != dp[nw][it][1])) {
            dp[msk][1] = dp[nw][it];
          }
        }
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int cmp = n + 1;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    ans -= a[i];
    p[i] = i;
  }
  a[n] = 0;
  p[n] = n;
  while (cmp > 1) {
    build_dp();
    for (int i = 0; i <= n; ++i) {
      int msk = ((1 << 18) - 1) ^ a[i];
      int me = get(i);
      if ((dp[msk][0][1] != -1) && (dp[msk][0][1] != me)) {
        array<long long, 2> cur = dp[msk][0];
        cur[0] += a[i];
        if (cur[0] > best[me][0]) {
          best[me] = cur;
        }
      } else if (dp[msk][1][1] != -1) {
        array<long long, 2> cur = dp[msk][1];
        cur[0] += a[i];
        if (cur[0] > best[me][0]) {
          best[me] = cur;
        }
      }
    }
    for (int i = 0; i <= n; ++i) {
      if (p[i] != i) {
        continue;
      }
      if (unite(i, best[i][1])) {
        --cmp;
        ans += best[i][0];
      }
    }
  }
  cout << ans << "\n";
  return 0;
}