#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const int maxm = maxn*10;
const int inf = 0x3f3f3f3f;
typedef long long ll;

int n,a[1<<18],b[1<<18],ans,cnt;
int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin>>n;
    for(int i = 0;i<(1<<n);++i) cin>>a[i];
    for(int i = 0;i<n;i++) 
        for(int j = 0;j<(1<<n);j++)
            if((j>>i)&1){
                if(a[j] > a[j^(1<<i)]) b[j] = max(b[j],a[j^(1<<i)]);
                else {
                    b[j] = max(a[j],b[j^(1<<i)]);
                    a[j] = a[j^(1<<i)];
                }
            }
    int ans = 0;
    for(int i = 1;i<(1<<n);i++) {
        ans = max(ans,a[i] + b[i]);
        printf("%d\n",ans);
    }
}
### Prompt

Your task is to create a Cpp solution to the following problem:

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.

Examples

Input

2
1 2 3 1


Output

3
4
5


Input

3
10 71 84 33 6 47 23 25


Output

81
94
155
155
155
155
155


Input

4
75 26 45 72 81 47 97 97 2 2 25 82 84 17 56 32


Output

101
120
147
156
156
178
194
194
194
194
194
194
194
194
194

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const int maxm = maxn*10;
const int inf = 0x3f3f3f3f;
typedef long long ll;

int n,a[1<<18],b[1<<18],ans,cnt;
int main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin>>n;
    for(int i = 0;i<(1<<n);++i) cin>>a[i];
    for(int i = 0;i<n;i++) 
        for(int j = 0;j<(1<<n);j++)
            if((j>>i)&1){
                if(a[j] > a[j^(1<<i)]) b[j] = max(b[j],a[j^(1<<i)]);
                else {
                    b[j] = max(a[j],b[j^(1<<i)]);
                    a[j] = a[j^(1<<i)];
                }
            }
    int ans = 0;
    for(int i = 1;i<(1<<n);i++) {
        ans = max(ans,a[i] + b[i]);
        printf("%d\n",ans);
    }
}
```