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 (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn main() {
    input!(n: usize, a: [u64; 1 << n]);
    let mut dp: Vec<_> = std::iter::successors(Some(0), |x| Some(x + 1))
        .map(|i| [i, i])
        .take(a.len())
        .collect();
    for bs in 1..1 << n {
        let mut v: Vec<_> = (0..n)
            .filter(|&i| bs >> i & 1 == 1)
            .map(|i| dp[bs ^ 1 << i].iter())
            .flatten()
            .copied()
            .chain(std::iter::once(bs))
            .collect();
        v.sort_by_key(|&i| std::cmp::Reverse(a[i]));
        v.dedup();
        dp[bs] = [v[0], v[1]];
    }
    let mut ans = 0;
    for bs in 1..1 << n {
        let x = a[dp[bs][0]] + a[dp[bs][1]];
        if ans < x {
            ans = x;
        }
        println!("{}", ans);
    }
}