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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include  <functional>
#include  <algorithm>
#include  <exception>
#include  <stdexcept>
#include  <streambuf>
#include  <iterator>
#include  <string.h>
#include  <stdlib.h>
#include  <typeinfo>
#include  <valarray>
#include  <iostream>
#include  <sstream>
#include  <istream>
#include  <stdio.h>
#include  <climits>
#include  <clocale>
#include  <complex>
#include  <csetjmp>
#include  <csignal>
#include  <cstdarg>
#include  <cstddef>
#include  <ctype.h>
#include  <cassert>
#include  <cstdlib>
#include  <utility>
#include  <cstring>
#include  <numeric>
#include  <ostream>
#include  <cwctype>
#include  <fstream>
#include  <iomanip>
#include  <math.h>
#include  <bitset>
#include  <cctype>
#include  <string>
#include  <vector>
#include  <limits>
#include  <locale>
#include  <memory>
#include  <cerrno>
#include  <time.h>
#include  <iosfwd>
#include  <cfloat>
#include  <cstdio>
#include  <cwchar>
#include  <cmath>
#include  <ctime>
#include  <deque>
#include  <queue>
#include  <stack>
#include  <list>
#include  <ios>
#include  <map>
#include  <set>
#include  <new>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ll long long
#define pii pair<int,int>
using namespace std;
const int maxn=300005;
int n,a[maxn];
int sum1[maxn],sum2[maxn];
int ans[maxn];
void upday(int val,int pos)
{
    if(val>sum1[pos])
    {
        sum2[pos]=sum1[pos];
        sum1[pos]=val;
    }
    else
    {
        if(val>sum2[pos])
        {
            sum2[pos]=val;
        }
    }
}
int main()
{
    cin>>n;
    for(int i=0;i<(1<<n);i++)
    {
        cin>>a[i];
        sum1[i]=a[i];
    }
    for(int i=0;i<(1<<n);i++)
    {
        for(int j=0;j<n;j++)
        {
            int t=1<<j;
            if(!(i&t))
            {
                int nowp=(i|t);
                int old=sum1[i];
                upday(old,nowp);
                old=sum2[i];
                upday(old,nowp);
            }
        }
    }
    for(int i=0;i<(1<<n);i++)
    {
        if(i!=0)
        ans[i]=max(ans[i-1],sum1[i]+sum2[i]);
        else
        ans[i]=sum1[i]+sum2[i];
    }
    for(int i=1;i<(1<<n);i++)
    {
        cout<<ans[i]<<endl;
    }
    return 0;
}