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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
int n,a[600000],vis[600000],mx[600000],mx2[600000],ans,inq[600000];
queue<int> q;
void upd(int x,int y) {
    if (mx[x]!=mx[y] and mx2[x]!=mx[y]) {
        if (a[mx[y]]>a[mx[x]]) {swap(mx[x],mx2[x]); mx[x]=mx[y];}
        else if (a[mx[y]]>a[mx2[x]]) mx2[x]=mx[y];
    }
    if (mx[x]!=mx2[y] and mx2[x]!=mx2[y] and a[mx2[y]]>a[mx2[x]]) mx2[x]=mx2[y];
}
int main() {
    cin>>n;
    for (int i=0; i<(1<<n); i++) cin>>a[i];
    for (int i=0; i<(1<<n); i++) {mx[i]=i; mx2[i]=(1<<n);}
    q.push(0);
    while (!q.empty()) {
        int cur=q.front();
        q.pop();
        if (vis[cur]) continue;
        vis[cur]=1;
        for (int i=0; i<n; i++) {
            if ((cur&(1<<i)) or vis[cur|(1<<i)]) continue;
            int nex=cur|(1<<i);
            if (!inq[nex]) q.push(nex);
            inq[nex]=1; upd(nex,cur);
        }
    }
    for (int i=1; i<(1<<n); i++) {
        ans=max(ans,a[mx[i]]+a[mx2[i]]);
        cout<<ans<<'\n';
    }
}