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)

// Date: 2020-02-25

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef long double LD;
typedef vector<int> VI;
typedef pair<LL, LL> pll;
typedef pair<int, int> pii;
#define FIO ios::sync_with_stdio(false);cin.tie(0)
#define rep(i, a, b) for(int i = int(a); i <= int(b); ++i)
#define per(i, b, a) for(int i = int(b); i >= int(a); --i)
#define mem(x, y) memset(x, y, sizeof(x))
#define all(x) (x).begin(),(x).end()
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const LL INF = 1e18;
const LL mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int N = 1e6 + 10;
LL qpow(LL x, LL y, LL MOD) {LL a=1; while(y){ if(y&1) a=a*x%MOD; x=x*x%MOD; y>>=1; } return a;}

LL S[N][2];
int a[N], b[N];

int main() {
    mem(S, -0x3f);
    int n; scanf("%d", &n);
    for(int i = 0; i < (1 << n); ++i) scanf("%d", &a[i]), S[i][0] = a[i];
    for(int k = 0; k < n; ++k) {
        for(int i = 0; i < (1 << n); ++i) {
            if(!((i >> k) & 1)) {
                int j = (i | (1 << k));
                b[0] = S[i][0], b[1] = S[i][1];
                b[2] = S[j][0], b[3] = S[j][1];
                sort(b, b+4);
                S[j][0] = b[3], S[j][1] = b[2];
            } 
        }
    }
    LL ans = 0;
    for(int i = 1; i < (1 << n); ++i) {
        ans = max(ans, S[i][0] + S[i][1]);
        printf("%lld\n", ans);
    }

    return 0;
}