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 = 'import sys\ninput = sys.stdin.readline\n\ndef I(): return int(input())\ndef MI(): return list(map(int, input().split()))\ndef LI(): return list(map(int, input().split()))\n\ndef main():\n    mod=10**9+7\n    N=I()\n    A=LI()\n    N2=pow(2,N)\n    \n    """\n    下位集合に対する高速ゼータ変換．\n    (Kがi,jを包含している）\n    ここでのgを適切なfの和として考えるのが基本だけど，\n    今回は和ではなくて上位2つの要素を持つ\n    """\n    \n    g=[[0,0]for _ in range(N2)]\n    \n    for i in range(N2):\n        g[i][0]=A[i]\n        \n    for k in range(N):#このkはKと違う，kbit目に着目する\n        for s in range(N2):#これが今回でいうK，\n            if ((s>>k)&1):#sのkbit目が立っていれば\n                #加法として，上位2つを選択する\n                L=[0,0,0,0]\n                L[0]=g[s][0]\n                L[1]=g[s][1]\n                L[2]=g[(s^(1<<k))][0]#sのkbit目をマスクしたgを持ってくる\n                L[3]=g[(s^(1<<k))][1]\n                L.sort()\n                g[s][0]=L[-1]\n                g[s][1]=L[-2]\n                #print(bin(s),k,g[s],L)\n                \n    ans=0\n    for k in range(1,N2):\n        ans=max(ans,sum(g[k]))\n        print(ans)\n        \n    \n                    \n                \n            \n        \n    \n    \n\nmain()\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14677