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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <algorithm>
// #include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#include <bitset>
#include <ctime>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef double db;
const int oo=2139063143;
const int N=1010000;
const int P=1000000007;
const db eps=1e-7;
#define pritnf printf
//char buf[1<<20|1],*p1=buf,*p2=buf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
template <typename T>
inline void sc (T &x)
{
    x=0; static int p; p=1; static char c; c=getchar();
    while (!isdigit(c)) { if (c=='-') p=-1; c=getchar(); }
    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }
    x*=p;
}
template <typename T>
inline void print (T x)
{
    if (x< 0) putchar('-'),x=-x;
    if (x>=10) print(x/10);
    putchar(x%10+'0');
}
template <typename T>
inline void pr (T x) { print(x),putchar('\n'); }
template <typename T1,typename T2>
void chkmax (T1 &A,T2 B) { A=A>=B?A:B; }
template <typename T1,typename T2>
void chkmin (T1 &A,T2 B) { A=A<=B?A:B; }
struct LY { int x,y; }a[N];
void update (int i,int d)
{
	if (a[i].y< d) a[i].y=d;
	if (a[i].y> a[i].x) swap (a[i].y,a[i].x);
}
int n;
int main ()
{
    // freopen (".in","r",stdin);
    // freopen (".out","w",stdout);
	sc(n);
	for (int i=0; i< 1<<n; i++)
	{
		int x; sc(x);
		a[i]=(LY){x,-oo};
	}
	for (int j=0; j< n; j++)
		for (int i=0; i< 1<<n; i++) if (!(i>>j&1))
			update (i|(1<<j),a[i].x),update (i|(1<<j),a[i].y);
	int ans=0;
	for (int i=1; i< 1<<n; i++)
	{
		chkmax (ans,a[i].x+a[i].y);
		pr(ans);
	}

    return 0;
}