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: Runtime Error)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define F first
#define S second
#define int long long
#define ll long long
//#define int  unsigned long long
#define pb push_back
//#define double long double

using namespace std;
using namespace __gnu_pbds;
typedef  tree< int , null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
const int mod = 1e9 + 7;
const int N = 200000;
set < int > ms;
int used[N], t[N], color[N], r[N], cv, v[N], s[N], u[N], n, m, val[N];
pair < int, int > p[N];
vector < pair < int, int > > g[N];
pair < int, int > ans[N];
int res[N];
main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    int n;
    cin >> n;
    for (int i = 0; i < (1 << n); i++){
        cin >> ans[i].F;
    }
    for (int j = 0; j < n; j++){
        for (int i = 0; i < (1 << n); i++){
            if (!((1 << j) & i)){
                int x = ans[i | (1 << j)].F;
                int y = ans[i | (1 << j)].S;
                int t = ans[i].F;
                int z = ans[i].S;
                multiset < int > s;
                s.insert(-x);
                s.insert(-y);
                s.insert(-t);
                s.insert(-z);
                auto it = s.begin();
                ans[i | (1 << j)].F = -(*it);
                it++;
                ans[i | (1 << j)].S = -(*it);
            }
        }
    }
    for (int i = 1; i < (1 << n); i++){
        res[i] = max(ans[i].F + ans[i].S, res[i - 1]);
        cout << res[i] << endl;

    }
}