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;
const int MOD=1e9+7;
//const int MOD=998244353;
const long long LINF=1e18;
#define int long long
//template
template <typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
/*
各bitがKより小さいi,jでA_i,A_jが一番大きいものをdp[K][0],dp[K][1]とする(i,jが入ってる)
答えは当然dp[K][0]+dp[K][1]
dp[K][0]とdp[K][1]の求め方は、Kの立ってるbitを1つ外した数字Lに対してのdp[L][0]とdp[L][1]、A[K]を見ればいいので
40個くらい見ればいいはず
*/
int dp[1234567][3];
//main
signed main(){
  int N;cin>>N;
  std::vector<int> v(1234567);
  for(int i=0;i<pow(2,N);i++)cin>>v[i];
  typedef pair<int,int> P;
  set<int> s;
  priority_queue<P> que;
  int mx=0;dp[0][0]=0;
  for(int i=1;i<pow(2,N);i++){
    s.clear();
    for(int j=0;j<N;j++){
      if((i>>j)&1){
        s.insert(dp[i^(1<<j)][0]);
        if((i^(1<<j)))s.insert(dp[i^(1<<j)][1]);
      }
    }
    //cout<<i<<" "<<s.size()<<endl;
    while(que.size())que.pop();//que.clear();
    for(int p:s)que.push(P(v[p],p));
    que.push(P(v[i],i));
    dp[i][0]=que.top().second;que.pop();
    dp[i][1]=que.top().second;que.pop();
    //cout<<i<<" "<<dp[i][0]<<" "<<dp[i][1]<<" "<<v[dp[i][0]]+v[dp[i][1]]<<endl;
    mx=max(mx,v[dp[i][0]]+v[dp[i][1]]);
    cout<<mx<<endl;
  }
}