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 <bits/stdc++.h>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
using namespace std;
using Int = long long;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<55)+1e9; // ~ 3.6 * 1e16
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
using T = tuple<Int,Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
ostream& operator<<(ostream& o,P p){return o<<"("<<p.first<<","<<p.second<<")";}
ostream& operator<<(ostream& o,T t){return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
istream& operator>>(istream& i,P &p){return i>>p.first>>p.second;}
ostream& operator<<(ostream& o,vector<auto> &a){Int i=0;for(auto t:a)o<<(i++?" ":"")<<t;return o;}
istream& operator>>(istream& i,vector<auto> &a){for(auto &t:a)i>>t;return i;}
void prArr(auto a,string s=" "){Int i=0;for(auto t:a)cout<<(i++?s:"")<<t;cout<<endl;}


signed main(){
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);
  
  Int n;
  cin>>n;
  const Int N = 1<<n;
  vector<Int> A(N);
  cin>>A;

  set<P> Top2[N];
  auto insert=[&](Int i,P t){
    Top2[i].insert(t);
    while(Top2[i].size() > 2) Top2[i].erase(Top2[i].begin());
  };
  
  for(Int i=1;i<N;i++){
    Int I = i;
    if(__builtin_popcount(i) == 1){
      I--;
      for(auto t:Top2[I]) insert(i, t);
    }
    insert(i ,P(A[I], I));
    for(Int j=0;j<n;j++){
      if(!(I>>j&1)) continue;
      Int k = I ^ (1<<j);
      insert(i, P(A[k],k));
      for(auto t:Top2[k]) insert(i, t);
    }
  }
  
  for(Int i=1;i<N;i++){
    Int ans = A[i] + A[0];
    if(Top2[i].size() >=2){
      Int a = (Top2[i].begin())->first;
      Int b = (++Top2[i].begin())->first;
      Max(ans, a + b);
    }
    cout<<ans<<endl;
  }
  return 0;
}