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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll read(){
    ll x=0,f=1;char c=getchar();
    while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0' && c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n;
int a[1000100];
pair<int,int> b[1000100];

void upd(int x,int y){
    int num1=b[x].first,num2=b[x].second,num3=b[y].first,num4=b[y].second;
    b[x].first=max(num1,num3);
    if(num1>num3) b[x].second=max(num2,num3);
    else b[x].second=max(num1,num4);
}

int main(){
    #ifdef LZT
    freopen("in","r",stdin);
    #endif
    n=read();
    for(int i=0;i<(1<<n);i++)
        a[i]=read();
    for(int i=0;i<(1<<n);i++)
        b[i].first=a[i],b[i].second=-1e9;
    for(int k=0;k<n;k++){
        //cout<<k<<endl;
        for(int i=0;i<(1<<n);i++){
            if((i&(1<<k))!=0) continue;
            upd(i|(1<<k),i);
            //cout<<(i|(1<<k))<<' '<<i<<endl;
        }
    }/*
    for(int i=0;i+1<(1<<n);i++){
        upd(i+1,i);
    }*/
    int lastans=0;
    for(int i=1;i<(1<<n);i++){
        lastans=max(lastans,b[i].first+b[i].second);
        printf("%d\n",lastans);
    }
    return 0;
}

/*
2
1 2 3 1
*/