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-----
2
1 2 3 1

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'from sys import stdin\nreadline = stdin.readline\n\nn = int(readline())\n\na = list(map(int, readline().split()))\nk_max = 1 << n\n\n# 和が最大 => 最大、2番目に大きいものの和\n# dp[k] == kを得るときの最大、2番目に大きいもの\n# == max(dp[i] | i はkの部分集合)\n# kの部分集合は k に1要素足りないものを列挙すれば十分。\n# (2要素以上足りないものは1要素足りないものの部分集合なので)\ndp = [[a[i], 0] for i in range(k_max)]\nfor i in range(n):\n    bit = (1 << i)  # i bit目に着目\n    for src in range(k_max):  # i bit目が0のものからibit目が1のものへ配るDP\n        if bit & src:\n            continue  # すでにi bit目が1なら無視\n        dst = bit | src\n        max_values = dp[dst] + dp[src]\n        max_values.sort()\n        dp[dst] = [max_values[-1], max_values[-2]]\n\n# k以下 => 前回の答え(k未満)とkに等しい時を合わせた結果\nans = 0\nfor k in range(1, k_max):\n    ans = max(ans, dp[k][0] + dp[k][1])\n    print(ans)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13347