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 <iostream>
#include <vector>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <stdlib.h>
#include <fstream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <iterator>
#include <iomanip>
#include <climits>
using namespace std;
#define ll long long int


int main(){
    ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
    
    ll n;
    cin >> n;
    ll N = 1 << n;
    vector<pair<ll, ll> > d(N);
    for(int i = 0; i < N; ++i) cin >> d[i].first;
    
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < N; ++j){
            if(j&(1<<i)){
                int s = j^(1<<i);
                if(d[j].first < d[s].first){
                    d[j].second = max(d[j].first, d[s].second);
                    d[j].first = d[s].first;
                }
                else d[j].second = max(d[j].second, d[s].first);
            }
        }
    }
    ll ans = 0;
    for(int i = 1; i < N; ++i){
        ans = max(ans, d[i].first + d[i].second);
        cout << ans << endl;
    }
}