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 = 300000, m = 1 << 18, mod = 998244353;
template <typename T>
void read(T &x) {
  int ch = getchar();
  x = 0;
  bool f = false;
  for (; ch < '0' || ch > '9'; ch = getchar()) f |= ch == '-';
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  if (f) x = -x;
}
void qmo(int &x) { x += (x >> 31) & mod; }
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) res = (long long)res * a % mod;
  return res;
}
template <typename T>
inline bool chmax(T &a, const T &b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <typename T>
inline bool chmin(T &a, const T &b) {
  if (a > b) return a = b, 1;
  return 0;
}
int n, a[N], fa[N];
long long ans;
int getfa(int x) {
  while (x != fa[x]) x = fa[x] = fa[fa[x]];
  return x;
}
void comb(int x, int y, int val) {
  if (a[x] && a[y]) {
    x = getfa(x);
    y = getfa(y);
    if (x != y) {
      ans += (a[x] + a[y] - 1ll) * val;
      fa[x] = y;
      a[y] = 1;
    }
  }
}
int main() {
  read(n);
  a[m] = 1;
  for (register int i = 0; i <= m; ++i) fa[i] = i;
  for (register int i = 1, x; i <= n; ++i) {
    read(x);
    ++a[x];
    ans -= x;
  }
  for (register int i = m - 1; ~i; --i) {
    for (register int j = i; j; j = (j - 1) & i) comb(j, i ^ j, i);
    comb(i, m, i);
    comb(0, i, i);
  }
  printf("%lld\n", ans);
}