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 = '\n"""\n\nhttps://atcoder.jp/contests/arc100/tasks/arc100_c\n\n情報を再利用するのがメイン\nK=Xの時は,K=X-1の時の答えに加え、 i or j == K の場合を考えればいい\nor がKとなる場合、0の桁は両方0で、1の桁は片方が1ならよい\n1では3通りの分け方があるので、3^N…\n\n大きいほうから考えてみる？\n最後は、当然最大のやつ2個→その2つのindex のORまではそれが保持\n\ndp[X] = X と orしてもXのままのindexの集合における最大 & 2番目\n→とすれば、1の桁全てに関して0にしてみた奴とXを見れば推移できる\n\nO(2^N * logMAX(A)) で解けそうだ\n\n"""\n\nfrom sys import stdin\n\nN = int(stdin.readline())\nA = list(map(int,stdin.readline().split()))\nA.append(float("-inf"))\n\ndp = []\nans = 0\naaa = []\n\nfor i in range(2**N):\n\n    if i == 0:\n        dp.append( ( 0,-1 ) )\n        continue\n\n    na,nb = i,-1\n    \n    for j in range(N):\n\n        if (2**j) & i > 0:\n\n            ta,tb = dp[i ^ (2**j)]\n            if A[na] < A[ta]:\n                nb = na\n                na = ta\n            elif A[nb] < A[ta] and ta != na:\n                nb = ta\n\n            if A[nb] < A[tb]:\n                nb = tb\n\n    #print (i,na,nb)\n    ans = max(A[na] + A[nb] , ans)\n    aaa.append(ans)\n    dp.append( ( na,nb ) )\n\nprint(("\\n".join(map(str,aaa))))\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13597