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 gc getchar()
inline int read(){
	int x=0,f=1;char c=gc;
	for(;c<48||c>57;c=gc)
		if(c=='-')f=-1;
	for(;c>47&&c<58;c=gc)
		x=x*10+c-48;
	return x*f;
}
#define io read()

#define maxn 1000005
#define maxm 10000005
int n,m,a[maxn];
int st[maxn],pv[maxm],p1[maxm],tot;
int in[maxn];
inline void side(int u,int v){	
	pv[++tot]=v;
	p1[tot]=st[u];
	st[u]=tot;
	++in[v];
}

inline void init(){
	n=io;m=1<<n;
	for(int i=0;i<m;++i)a[i]=io;
	for(int i=0;i<m;++i){
		for(int b=1;b<=i;b<<=1){
			if(i&b)side(i,i^b);
		}
	}
}

int q[maxn],l,r;
int dp[maxn],ans;
struct pi{
	int x,y;
	pi(int xx=0,int yy=0):x(xx),y(yy){}
};
pi mx[maxn];

int b[100];
inline bool cmp(int x,int y){return a[x]>a[y]||a[x]==a[y]&&x>y;}

inline void solve(){
	l=1;r=0;
	for(int i=0;i<m;++i)
		if(!in[i])q[++r]=i;
	while(l<=r){
		int x=q[l++];
		for(int i=st[x];i;i=p1[i]){
			int v=pv[i];
			if(!--in[v])q[++r]=v;
		}
	}
	
	a[m]=dp[m]=0;
	for(int i=r;i;--i){
		int x=q[i];
		int cnt=0;
		b[++cnt]=x;
		b[++cnt]=m;
		for(int j=st[x];j;j=p1[j]){
			int v=pv[j];
			b[++cnt]=mx[v].x;
			b[++cnt]=mx[v].y;
		}
		sort(b+1,b+cnt+1,cmp);
		cnt=unique(b+1,b+cnt+1)-b-1;
		mx[x]=pi(b[1],b[2]);
		dp[x]=a[b[1]]+a[b[2]];
	}
	
	for(int i=1;i<m;++i)dp[i]=max(dp[i-1],dp[i]);
	for(int i=1;i<m;++i)printf("%d\n",dp[i]);
}

int main(){
	init();
	solve();
	return 0;
}