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>
struct MI {
 private:
  char bb[4096];
  FILE *f;
  char *bs, *be;
  char e;
  bool o, l;

 public:
  MI() : f(stdin) {}
  inline char get() {
    if (o) {
      o = 0;
      return e;
    }
    if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
    if (bs == be) {
      l = 1;
      return -1;
    };
    return *bs++;
  }
  inline void unget(char c) {
    o = 1;
    e = c;
  }
  template <typename T>
  inline T read() {
    T r;
    *this > r;
    return r;
  }
  template <typename T>
  inline MI &operator>(T &);
};
template <typename T>
struct Q {
  const static bool U = T(-1) >= T(0);
  inline void operator()(MI &t, T &r) const {
    r = 0;
    char c;
    bool y = 0;
    if (U)
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (isdigit(c)) break;
      }
    else
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (c == '-') {
          c = t.get();
          if (isdigit(c)) {
            y = 1;
            break;
          };
        } else if (isdigit(c))
          break;
        ;
      };
    for (;;) {
      if (c == -1) goto E;
      if (isdigit(c))
        r = r * 10 + (c ^ 48);
      else
        break;
      c = t.get();
    }
    t.unget(c);
  E:;
    if (y) r = -r;
  }
};
template <>
struct Q<char> {};
template <typename T>
inline MI &MI::operator>(T &t) {
  Q<T>()(*this, t);
  return *this;
}
template <typename T>
std::ostream &operator<(std::ostream &out, const T &t) {
  return out << t;
}
using std::cout;
MI cin;
const int nmax = 262170;
template <typename T>
inline bool getmax(T &a, const T &b) {
  return a < b && (a = b, true);
}
int n, cnt[nmax], f[nmax], rem;
long long ans;
int getf(int x) { return f[x] == x ? x : (f[x] = getf(f[x])); }
inline void merge(int x, int y) {
  assert(!(x & y));
  if (!cnt[x] || !cnt[y]) return;
  const int w(x | y);
  x = getf(x);
  y = getf(y);
  if (x == y) return;
  ans += (long long)w * (cnt[x] + cnt[y] - 1);
  f[x] = y;
  cnt[y] = 1;
  if (!--rem) {
    cout < ans < ('\n');
    exit(0);
  }
}
int main() {
  cin > n;
  int mx = 0;
  cnt[0] = 1;
  for (int i = 1; i <= n; ++i) {
    const int x((cin.read<int>()));
    rem += !cnt[x]++;
    ans -= x;
    f[x] = x;
    getmax(mx, x);
  }
  mx = 1 << (32 - __builtin_clz(mx));
  for (int i = mx - 1; ~i; --i) {
    for (int j = i; j; j = i & (j - 1)) {
      if (j < (i ^ j)) break;
      merge(j, i ^ j);
    }
  }
  cout < ans < ('\n');
}