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)

#include <iostream>
#include <algorithm>
#include <vector>

using Int = long long int;
template <class T>
using Vector = std::vector<T>;

struct G {
    Vector<Int> idx;
    static Vector<Int> a;

    G& operator+=(const G& opp) {
        std::copy(opp.idx.begin(), opp.idx.end(),
                  std::back_inserter(idx));
        std::sort(idx.begin(), idx.end(), [&](Int i, Int j) {
            return a[i] == a[j] ? i < j : a[i] > a[j];
        });
        idx.erase(std::unique(idx.begin(), idx.end()), idx.end());
        idx.resize(2);
        return *this;
    }
};
Vector<Int> G::a;

int main() {
    Int n;
    std::cin >> n;
    Vector<Int> a(1 << n);
    for (auto& x : a) std::cin >> x;
    G::a = a;

    Vector<G> g(1 << n);
    for (Int s = 0; s < (1 << n); ++s) {
        g[s].idx.push_back(s);
    }

    // 高速ゼータ変換 下位集合版
    for (Int k = 0; k < n; ++k) {
        for (Int b = 0; b < (1 << n); ++b) {
            if ((b >> k) & 1) {
                g[b] += g[b ^ (1 << k)];
            }
        }
    }

    Int max = 0;
    for (Int s = 1; s < (1 << n); ++s) {
        max = std::max(max, a[g[s].idx[0]] + a[g[s].idx[1]]);
        std::cout << max << std::endl;
    }
    return 0;
}