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>
#pragma optimization_level 3
using namespace std;
long double TL = 2.0;
bool is_tl = 0;
int CALLS = 0;
inline bool IS() {
  if (++CALLS == 1000) {
    CALLS = 0;
    is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
  }
  return is_tl;
}
template <typename T1, typename T2>
inline void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T>
inline void print(T &a) {
  for (auto &U : a) cout << U << ' ';
  cout << '\n';
}
const int N = (1 << 18), NS = 1e3 + 7, lg = 20, sq = 550, inf = 1e9 + 7,
          mod = 998244353, SM = 1e2 + 7;
const long double eps = 1e-9;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
int a[N], p[N], cnt[N];
int get(int x) {
  if (p[x] < 0) return x;
  return p[x] = get(p[x]);
}
int unite(int x, int y) {
  x = get(x), y = get(y);
  if (x == y) return 0;
  if (p[x] > p[y]) swap(x, y);
  int res = (p[x] == -1 ? cnt[x] : 1) + (p[y] == -1 ? cnt[y] : 1) - 1;
  p[x] += p[y];
  p[y] = x;
  return res;
}
void solve() {
  int n;
  long long ans = 0;
  fill(p, p + N, -1);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
    ans -= a[i];
  }
  cnt[0]++;
  for (int i = N - 1; i >= 0; i--) {
    for (int j = i; j > 0; j = ((j - 1) & i)) {
      int nxt = (i ^ j);
      if (cnt[j] && cnt[nxt]) {
        ans += 1ll * i * unite(j, nxt);
      }
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}