Problem:
Score : 700 points

Problem Statement

There is an integer sequence of length 2^N: A_0, A_1, ..., A_{2^N-1}. (Note that the sequence is 0-indexed.)

For every integer K satisfying 1 \leq K \leq 2^N-1, solve the following problem:

Let i and j be integers. Find the maximum value of A_i + A_j where 0 \leq i < j \leq 2^N-1 and (i or j) \leq K.
Here, or denotes the bitwise OR.

Constraints

1 \leq N \leq 18

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 ... A_{2^N-1}

Output

Print 2^N-1 lines.
In the i-th line, print the answer of the problem above for K=i.

Sample Input 1

2
1 2 3 1

Sample Output 1

3
4
5

For K=1, the only possible pair of i and j is (i,j)=(0,1), so the answer is A_0+A_1=1+2=3.

For K=2, the possible pairs of i and j are (i,j)=(0,1),(0,2).
When (i,j)=(0,2), A_i+A_j=1+3=4. This is the maximum value, so the answer is 4.

For K=3, the possible pairs of i and j are (i,j)=(0,1),(0,2),(0,3),(1,2),(1,3),(2,3) .
When (i,j)=(1,2), A_i+A_j=2+3=5. This is the maximum value, so the answer is 5.

Sample Input 2

3
10 71 84 33 6 47 23 25

Sample Output 2

81
94
155
155
155
155
155

Sample Input 3

4
75 26 45 72 81 47 97 97 2 2 25 82 84 17 56 32

Sample Output 3

101
120
147
156
156
178
194
194
194
194
194
194
194
194
194

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#ifndef LOCAL
#pragma GCC optimize ("O3")
#endif

#include <bits/stdc++.h>

using namespace std;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return {i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (c it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define imie(x...) " [" #x ": " << (x) << "] "

using ld = long double;
using ll = long long;

constexpr int mod = 1000 * 1000 * 1000 + 7;
constexpr int odw2 = (mod + 1) / 2;

void OdejmijOd(int& a, int b) { a -= b; if (a < 0) a += mod; }
int Odejmij(int a, int b) { OdejmijOd(a, b); return a; }
void DodajDo(int& a, int b) { a += b; if (a >= mod) a -= mod; }
int Dodaj(int a, int b) { DodajDo(a, b); return a; }
int Mnoz(int a, int b) { return (ll) a * b % mod; }
void MnozDo(int& a, int b) { a = Mnoz(a, b); }
int Pot(int a, int b) { int res = 1; while (b) { if (b % 2 == 1) MnozDo(res, a); a = Mnoz(a, a); b /= 2; } return res; }
int Odw(int a) { return Pot(a, mod - 2); }
void PodzielDo(int& a, int b) { MnozDo(a, Odw(b)); }
int Podziel(int a, int b) { return Mnoz(a, Odw(b)); }
int Moduluj(ll x) { x %= mod; if (x < 0) x += mod; return x; }

template <typename T> T Maxi(T& a, T b) { return a = max(a, b); }
template <typename T> T Mini(T& a, T b) { return a = min(a, b); }

struct Wyn {
  int x, y;
  int vx, vy;

  Wyn() : x(0), y(0), vx(-1), vy(-1) {}

  void Dodaj(int val, int v) {
    if (v == vx or v == vy) return;
    if (val > x) {
      y = x;
      vy = vx;
      x = val;
      vx = v;
    } else if (val > y) {
      y = val;
      vy = v;
    }
  }

  void Polacz(const Wyn& w) {
    Dodaj(w.x, w.vx);
    Dodaj(w.y, w.vy);
  }

  int Oblicz() const {
    return x + y;
  }
};

int n;
int A[1 << 18];
Wyn dp[1 << 18];

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (1 << n); i++) {
    cin >> A[i];

    Wyn& w = dp[i];
    w.Dodaj(A[i], i);

    for (int j = 0; j < n; j++) {
      if ((i >> j) & 1) {
        w.Polacz(dp[i ^ (1 << j)]);
      }
    }
    debug() << imie(i) imie(w.x) imie(w.vx) imie(w.y) imie(w.vy);
  }

  for (int i = 1; i < (1 << n); i++) {
    int odp = dp[i].Oblicz();;
    for (int j = 0; j < n; j++) {
      if ((i >> j) & 1) {
        Maxi(odp, dp[(i ^ (1 << j)) | ((1 << j) - 1)].Oblicz());
      }
    }
    cout << odp << '\n';
  }
  return 0;
}