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: Wrong Answer)

#include<iostream>
#include<set>
#include<vector>
using namespace std;
typedef long long li;
#define chmax(a,b) (a<(b)?a=(b),1:0)
#define repa(i,a,n) for(int i=(a);i<(n);i++)
#define rep(i,n) for(li i=0;i<(n);i++)
#define df 0
template<class T> void print(const T& t){ cout << t << "\n"; }
template<class T, class... Ts> void print(const T& t, const Ts&... ts) { cout << t; if (sizeof...(ts)) cout << " "; print(ts...); }
// Container コンテナ型, map以外
template< template<class ...> class Ctn,class T>
std::ostream& operator<<(std::ostream& os,const Ctn<T>& v){
  auto itr=v.begin();
  while(itr!=v.end()){
    if(itr!=v.begin())cout << " ";
    cout << *(itr++);
  }
  return os;
}

int ord(li bs){
  li b=bs;int i=1;
  while(b){
    b>>i;
    bs|=b;
    i*=2;
  }
  return __builtin_popcount(bs);
}

int _2beki(li bs){
  return bs&(bs-1)==0;
}

struct data{
  li a;
  li itr;
  data(li x=0,li i=-1){a=x;itr=i;}
  int operator<(const data& d) const
  {
    if(a==d.a)return itr<d.itr;
    return a>d.a;
  }
  li operator+(data d){
    return a+d.a;
  }
};
std::ostream& operator<<(std::ostream& os,const data& d){
  printf("[%d]%ld",d.itr,d.a);
  return os;
}

int main(){
  int n;  cin >>n;
  li lim=1<<n;
  vector<data> a(lim);
  rep(i,lim){
    int x; cin >>x;
    a[i]={x,i};
  }
  if(df)print(a);
  li ans=0;
  vector<data> fst(lim),snd(lim);
  fst[0]=snd[0]=a[0];
  repa(i,1,lim){
    set<data> s;
    s.insert(a[i]);
    li j=1<<__builtin_ctzll(i);
    if(df)print(i,j,i-j);
    s.insert(fst[i-j]);
    s.insert(snd[i-j]);
    s.insert(a[j]);
    auto itr=s.begin();
    fst[i]=*itr;
    itr++;
    snd[i]=*itr;
    chmax(ans,fst[i]+snd[i]);
    print(ans);
  }
  if(df){
    print(fst);
    print(snd);
  }
}