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 LuaJIT (2.0.4) (judge verdict: Time Limit Exceeded)

local function bdp_prepare(n)
  local tot = 1
  local alltask = {}
  local stagetask = {}
  for i = 1, n do
    tot = tot * 2
    stagetask[i] = {}
  end
  local a = {}
  local a0 = io.read("*n")
  for i = 1, tot - 1 do
    a[i] = io.read("*n")
    if a[i] < a0 then
      alltask[i] = {tot, i}
    else
      alltask[i] = {i, tot}
    end
  end
  a[tot] = a0
  for i = 1, tot - 1 do
    local ti = i
    local cnt = 0
    for j = 1, n do
      if ti % 2 == 1 then
        cnt = cnt + 1
        ti = ti - 1
      end
      ti = ti / 2
    end
    table.insert(stagetask[cnt], i)
  end
  return tot, a, alltask, stagetask
end

local function bdp_walk(a, alltask, src, dst)
  local dst1 = alltask[dst][1]
  local dst2 = alltask[dst][2]
  local src1 = alltask[src][1]
  local src2 = alltask[src][2]
  local tmp = {dst1, dst2, src1, src2}
  table.sort(tmp, function(x, y) return a[x] > a[y] end)
  for i = 4, 2, -1 do
    if tmp[i] == tmp[i - 1] then
      table.remove(tmp, i)
    end
  end
  alltask[dst][1] = tmp[1]
  alltask[dst][2] = tmp[2]
end

local function bdp_doall(n, a, alltask, stagetask)
  for stage = 1, n - 1 do
    local stlen = #stagetask[stage]
    for i_stagetask = 1, stlen do
      local used = stagetask[stage][i_stagetask]
      local mul = 1
      local tmp = used
      for j = 1, n do
        if tmp % 2 == 0 then
          bdp_walk(a, alltask, used, used + mul)
        else
          tmp = tmp - 1
        end
        tmp = tmp / 2
        mul = mul * 2
      end
    end
  end
end

local n = io.read("*n")
local tot, a, alltask, stagetask = bdp_prepare(n)
bdp_doall(n, a, alltask, stagetask)
local prv = 0
for i = 1, tot - 1 do
  local cur = a[alltask[i][1]] + a[alltask[i][2]]
  if cur < prv then cur = prv end
  print(cur)
  prv = cur
end