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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class Main {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    PriorityQueue<Pair> maxQ = new PriorityQueue<>(Collections.reverseOrder());
    PriorityQueue<Pair> minQ = new PriorityQueue<>();
    int qSize = 8;

    int n = Integer.parseInt(br.readLine());
    String[] s = br.readLine().split(" ");
    long[] arr = new long[n];
    for (int i=0; i<(1<<n); i++) {
      long v = Long.parseLong(s[i]);
      Pair p = new Pair(i,v);
      if (maxQ.size()<qSize) {
        maxQ.offer(p);
        minQ.offer(p);
      } else {
        if (v>minQ.peek().val) {
          Pair r = minQ.poll();
          maxQ.remove(r);
          maxQ.offer(p);
          minQ.offer(p);
        }
      }
      if (i>0) {
        int index1 = 0;
        long bestSum = -1;
        for (Pair p1:maxQ) {
          int index2 = 0;
          for (Pair p2:maxQ) {
            if (p1.index!=p2.index && (p1.index|p2.index)<=i  && p1.val+p2.val>bestSum) {
              bestSum = p1.val+p2.val;
            }
            index2++;
            if (index2>=index1)
              break;
              
          }
          index1++;
        }
        if (bestSum==-1) {
          throw new NullPointerException();
        }
        sb.append(bestSum);
        sb.append("\n");
      }
    }
    System.out.print(sb);
  }
}
class Pair implements Comparable<Pair> {
  int index;
  long val;
  public Pair(int ii, long vv) {
    index = ii;
    val = vv;
  }
  public int compareTo(Pair p2) {
    return (int)(val-p2.val);
  }
}