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 <utility>
#define llint long long

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[1<<18];
P p[1<<18];
llint dp[1<<18];

P maxp(P x, P y)
{
	if(x.first == y.first) return make_pair(x.first, y.first);
	if(y > x) swap(x, y);
	if(x.second > y.first) return x;
	else return make_pair(x.first, y.first);
}

int main(void)
{
	cin >> n;
	llint S = 1<<n;
	
	for(int i = 0; i < S; i++) cin >> a[i];
	for(int i = 0; i < S; i++) p[i] = make_pair(a[i], 0);
	
	for(int i = 0; i < n; i++){
		for(int j = 0; j < S; j++){
			if(j&(1<<i)) p[j] = maxp(p[j], p[j^(1<<i)]);
		}
	}
	
	dp[0] = 0;
	for(int i = 1; i < S; i++){
		dp[i] = max(dp[i-1], p[i].first + p[i].second);
	}
	
	for(int i = 1; i < S; i++) cout << dp[i] << endl;
	return 0;
}