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 = 4e5 + 7;
int n, a[N], cnt[N], lim = 0, mx = 0, fa[N];
long long ans = 0;
inline int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
inline void merge(int u, int v, long long w) {
  if (cnt[u] && cnt[v]) {
    u = find(u), v = find(v);
    if (u != v) {
      ans += 1ll * w * (cnt[v] + cnt[u] - 1);
      fa[u] = v;
      cnt[v] = 1;
    }
  }
}
int main() {
  scanf("%d", &n);
  cnt[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
    mx = max(mx, a[i]);
    ans -= a[i];
  }
  lim = 1 << 32 - __builtin_clz(mx);
  for (int i = 0; i <= lim - 1; i++) fa[i] = i;
  for (int i = lim - 1; ~i; i--) {
    for (int j = i; j; j = (j - 1) & i) {
      if (cnt[j] && cnt[i ^ j]) merge(j, i ^ j, i);
    }
    merge(i, 0, i);
  }
  printf("%lld\n", ans);
  return 0;
}