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)

fn main() {
    let n = readln();
    let a: Vec<usize> = readln();
    let l = (1<<n) as usize;
    let mut dp = vec![BinaryHeap::new(); l];
    for s in 0..l {
        dp[s].push(Rev((a[s], s)));
    }
    for s in 0..l {
        for i in 0..n {
            if (s >> i & 1) == 1 {
                let ns = s ^ (1<<i);
                let values: Vec<Rev<(usize, usize)>> = dp[ns].iter().cloned().collect();
                for value in values {
                    dp[s].push(value);
                }
            }
        }
        while dp[s].len() > 2 {
            dp[s].pop();
        }
    }
    let mut cur = 0;
    for s in 1..l {
        let a = (dp[s].pop().unwrap().0).0;
        let b = (dp[s].pop().unwrap().0).0;
        cur = max(cur, a + b);
        println!("{}", cur);
    }
}

use std::cmp::Ordering;
#[derive(Eq, PartialEq, Clone, Debug)]
#[doc = " Equivalent to std::cmp::Reverse"]
pub struct Rev<T>(pub T);
impl<T: PartialOrd> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Rev<T>) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
}
impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Rev<T>) -> Ordering {
        other.0.cmp(&self.0)
    }
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, VecDeque, BinaryHeap};

pub trait FromLn {
    fn fromln(s: &str) -> Self;
}
pub fn readln<T: FromLn>() -> T {
    let mut buf = String::new();
    let _ = ::std::io::stdin().read_line(&mut buf).unwrap();
    T::fromln(buf.trim())
}
pub fn readlns<T: FromLn>(n: usize) -> Vec<T> {
    let mut vs = vec![];
    for _ in 0..n {
        vs.push(readln());
    }
    vs
}
macro_rules! fromln_primitives {
    ($($t:ty),*) => { $(
        impl FromLn for $t {
            fn fromln(s: &str) -> $t {
                s.parse().unwrap()
            }
        }
    )* }
}
fromln_primitives!(
    String,
    bool,
    f32,
    f64,
    isize,
    i8,
    i16,
    i32,
    i64,
    usize,
    u8,
    u16,
    u32,
    u64
);
impl<T> FromLn for Vec<T>
where
    T: FromLn,
{
    fn fromln(s: &str) -> Vec<T> {
        s.split_whitespace().map(T::fromln).collect()
    }
}
impl FromLn for Vec<char> {
    fn fromln(s: &str) -> Vec<char> {
        s.chars().collect()
    }
}
macro_rules! fromln_tuple {
    ($($t:ident),*) => {
        impl<$($t),*> FromLn for ($($t),*) where $($t: FromLn),* {
            fn fromln(s: &str) -> ($($t),*) {
                let mut it = s.split_whitespace();
                let t = ($($t::fromln(it.next().unwrap())),*);
                assert_eq!(it.next(), None);
                t
            }
        }
    }
}
fromln_tuple!(A, B);
fromln_tuple!(A, B, C);
fromln_tuple!(A, B, C, D);
fromln_tuple!(A, B, C, D, E);
fromln_tuple!(A, B, C, D, E, F);