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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
int power(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y / 2, x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
const int N = (1 << 18);
pair<int, int> ar[N];
int freq[N];
void ini() {
  int i;
  for (i = 0; i < N; i++) ar[i].first = i, ar[i].second = 1;
}
int root(int x) {
  if (ar[x].first != x) ar[x].first = root(ar[x].first);
  return ar[x].first;
}
int uni(int x, int y) {
  int p = root(x);
  int q = root(y);
  if (p == q) return 0;
  int res = freq[x] + freq[y] - 1;
  if (ar[p].second > ar[q].second) {
    ar[q].first = ar[p].first;
  } else if (ar[p].second < ar[q].second) {
    ar[p].first = ar[q].first;
  } else {
    ar[p].first = ar[q].first;
    ar[q].second += ar[p].second;
  }
  freq[x] = 1;
  freq[y] = 1;
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ini();
  int n;
  cin >> n;
  long long ans = 0;
  for (int i = 0; i < (n); ++i) {
    int x;
    cin >> x;
    freq[x]++;
    ans -= x;
  }
  freq[0]++;
  for (int i = N - 1; i > 0; i--) {
    for (int j = i; j > 0; j = (j - 1) & i) {
      if (freq[j] and freq[i ^ j]) ans += 1LL * i * uni(j, i ^ j);
    }
  }
  cout << ans;
}