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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}

import qualified Data.ByteString.Lazy.Char8 as B
import Data.Maybe
import Data.Bits
import qualified Data.IntMap.Strict as IM
import Data.List
import Debug.Trace

bsReadInts :: B.ByteString -> [Int]
bsReadInts = map (fst . fromJust . B.readInt) . B.words

cum_max :: [Int] -> [Int]
cum_max = tail . scanl' max 0

nextZeta :: Int -> IM.IntMap (Int,Int) -> Int -> IM.IntMap (Int,Int)
nextZeta en im i = foldl' op im [0..en-1]
  where
    op !im !x | testBit x i = IM.adjust (f (im IM.! (clearBit x i))) x im
              | otherwise   = im
      where
{-
        f (q1,q2) (p1,p2) = (r1,r2)
          where r1:r2:_ = sortOn (\p -> -p) [p1,p2,q1,q2]
-}
        f !x !y = let (q1,q2) = x
                      (p1,p2) = y
                  in
                    if p1 >= q1 then
                      if p2 >= q1 then (p1,p2)
                      else (p1,q1)
                    else
                      if q2 >= p1 then (q1,q2)
                      else (q1,p1)

zeta :: Int -> [Int] -> IM.IntMap (Int, Int)
zeta n as = foldl' (nextZeta en) initmap [0..n-1] 
  where
    en = 2^n
    initmap = IM.fromList (zip [0..en-1] [(a,0) | a <- as])

solve :: Int -> [Int] -> [Int]
solve n as = cum_max $ tail $
                  map ((\(p,q)->p+q) . snd) $ IM.toAscList $ zeta n as

main :: IO ()
main = do
  conts <- B.getContents
  let n : as = bsReadInts conts
      answer = solve n as
  putStr (unlines (map show answer))