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;
vector<vector<int> > v(200001);
int parent[200001];
inline int Find(int i) {
  return i == parent[i] ? i : parent[i] = Find(parent[i]);
}
inline int Union(int i, int j) {
  i = Find(i), j = Find(j);
  if (i == j) {
    return 0;
  }
  if (i > j) {
    swap(i, j);
  }
  parent[j] = i;
  return 1;
}
inline int FormEdge(const int i, const int j) {
  return Union(v[i].back(), v[j].back());
}
inline int CollectAns(const int i, const int j) {
  if (i >= 200001 || j >= 200001 || v[i].empty() || v[j].empty()) {
    return 0ll;
  }
  int ans = 0;
  for (; v[i].size() > 1; v[i].pop_back()) {
    ans += FormEdge(i, j);
  }
  for (; v[j].size() > 1; v[j].pop_back()) {
    ans += FormEdge(i, j);
  }
  ans += FormEdge(i, j);
  return ans;
}
inline long long CollectAns(const int k) {
  int ans = 0;
  for (int i = k; i > 0; i = (i - 1) & k) {
    int j = k ^ i;
    if (j > i) {
      continue;
    }
    ans += CollectAns(i, j);
  }
  return 1ll * k * ans;
}
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    v[x].push_back(i);
    parent[i] = i;
    ans -= x;
  }
  v[0].push_back(0);
  for (int k = ((1 << 18) - 1); k > -1; --k) {
    ans += CollectAns(k);
  }
  printf("%lld\n", ans);
  return 0;
}