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 3.8.0) (judge verdict: Accepted)

#include <unistd.h>

char ibuf[(11<<18)+5];
char *ibufe = ibuf-1;

extern inline void readall(){
  int k, t = 0;
  while((k=read(STDIN_FILENO, ibuf+t, sizeof(ibuf)-t))>0) t += k;
}

int read_uint(){
  int x=0;
  while(*(++ibufe) <'0');
  do {
    x *= 10;
    x += *ibufe-'0';
  } while(*(++ibufe) >='0');

  return x;
}

char buf[40];
char *bufe = buf;

void write_uintln(int x){
  int i;
  static char tmp[13];
  if(x==0){
    *bufe++ = '0';
    *bufe++ = '\n';
    return;
  }

  for(i=0; x; i++){
    tmp[i] = '0' + x % 10;
    x /= 10;
  }
  for(i--; i >= 0; i--){
    *bufe++ = tmp[i];
  }
  *bufe++ = '\n';
}

extern inline void writeall(){
  int k, t = 0;
  while((k=write(STDOUT_FILENO, buf+t, bufe-buf-t))>0) t += k;
}


#define MAX(x, y) ((x) > (y) ? (x) : (y))

int n;
int A[1<<18];
int B[1<<18];

void zeta(int n){
  int i, j;
  int m = 1 << n;
  for(i=1;i<m;i<<=1){
    for(j=0;j<m;j++){
      if(j&i){
        if(A[j] >= A[j^i]){
          if(A[j^i] > B[j]) B[j] = A[j^i];
        }
        else {
          B[j] = MAX(A[j], B[j^i]);
          A[j] = A[j^i];
        }
      }
    }
  }
}


int main(){
  int i, max;
  readall();
  n = read_uint();
  for(i=0;i<(1<<n);i++){
    A[i] = read_uint();
  }
  zeta(n);
  max = 0;
  for(i=1;i<(1<<n);i++){
    max = MAX(max, A[i]+B[i]);
    write_uintln(max);
  }
  writeall();
  return 0;
}