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 <bits/stdc++.h>
using namespace std;
#define MAXN 262144
#define rint register int
inline int rf(){int r;int s=0,c;for(;!isdigit(c=getchar());s=c);for(r=c^48;isdigit(c=getchar());(r*=10)+=c^48);return s^45?r:-r;}
int A[MAXN+5], Ans[MAXN+5], n, k; vector<int> R[MAXN+5]; inline bool Amax(int i, int j){return A[i]^A[j]?A[i]>A[j]:i<j;}
void Merge(int a, int b){for(auto v:R[b]) R[a].push_back(v); sort(R[a].begin(),R[a].end(),Amax); R[a].erase(unique(R[a].begin(),R[a].end()),R[a].end()); R[a].resize(min((int)R[a].size(),2));}
int main()
{
	k = rf(); n = 1<<k; generate(A,A+n,rf); for(rint i = 0; i < n; R[i].push_back(i), i++); for(rint i = 0, j; i < n; i++) for(j = 1; j < n; i&j ? Merge(i,i^j),0 : 0, j <<= 1);
	for(rint i = 1; i < n; Ans[i] = Ans[i-1], R[i].size()>1?Ans[i]=max(Ans[i],A[R[i][0]]+A[R[i][1]]):0, printf("%d\n",Ans[i]), i++); return 0;
}