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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n = read::<usize>();
    let m = 1 << n;
    let mut aseq = vec![0; m];
    for i in 0..m {
        aseq[i] = read::<i64>();
    }
    let mut anss = vec![vec![(0, 0), (0, 0)]; m];
    for i in 1..m {
        let mut candidates = vec![(aseq[i], i), (aseq[0], 0)];
        candidates.push(anss[i - 1][0]);
        candidates.push(anss[i - 1][1]);
        for j in 0..n {
            if (i & (1 << j)) == 0 {
                continue;
            }
            let t = i ^ (1 << j);
            candidates.push(anss[t][0]);
            candidates.push(anss[t][1]);
        }
        candidates.sort();
        candidates.dedup();
        candidates.reverse();
        anss[i][0] = (aseq[i], i);
        anss[i][1] = (aseq[0], 0);
        for j in 0..candidates.len() {
            for k in j + 1..candidates.len() {
                if i < (candidates[j].1 | candidates[k].1) {
                    continue;
                }
                if anss[i][0].0 + anss[i][1].0 < candidates[j].0 + candidates[k].0 {
                    anss[i][0] = candidates[j];
                    anss[i][1] = candidates[k];
                }
            }
        }
        // println!("{:?}", candidates);
        // println!("{:?}", anss[i]);
    }
    for i in 1..m {
        println!("{}", anss[i][0].0 + anss[i][1].0);
    }
}