The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.

The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.

One move is choosing an integer i (1 ≤ i ≤ n) such that ai > 0 and an integer t (t ≥ 0) such that i + 2t ≤ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).

You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 ≤ k < n).

Input

The first input line contains a single integer n. The second line contains n integers ai (0 ≤ ai ≤ 104), separated by single spaces.

The input limitations for getting 20 points are: 

  * 1 ≤ n ≤ 300



The input limitations for getting 50 points are: 

  * 1 ≤ n ≤ 2000



The input limitations for getting 100 points are: 

  * 1 ≤ n ≤ 105

Output

Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams, or the %I64d specifier.

Examples

Input

4
1 0 1 2


Output

1
1
3


Input

8
1 2 3 4 5 6 7 8


Output

1
3
6
10
16
24
40
#include <iostream>
#include <cmath>
#include <stdio.h>
using namespace std;

typedef struct NODE
{
    int i;
    int k;
    int step;
}ABC;
ABC A[10000000];

int n;

int pp(int x,int y)
{
    int i=0;
    int min=1;
    for(i=0;i<y;i++)
    {
        min=min*x;
    }
    return min;
}

int bfs(int i,int k, int step)
{
    int t=0,j;
    int min=99999999;
    /*if(i>k) {return step;}
    for(t=0;i+pp(2,t)<=n;t++)
    {
        flag=dfs(i+pp(2,t),k,step+1);
        if(flag<min)
        {
            min = flag;
        }
    }
    return min;*/
    A[0].i=i;
    A[0].k=k;
    A[0].step=step;
    i=0;
    j=1;
    while(A[i].i<=A[i].k)
    {
        for(t=0;A[i].i+pp(2,t)<=n;t++)
        {
            A[j].i=A[i].i+pp(2,t);
            A[j].k=A[i].k;
            A[j].step=A[i].step+1;
            j++;
            if(j>=10000000) j=j%10000000;
        }
        i++;
        if(i>=10000000) i=i%10000000;

    }
    return A[i].step;
}



int main()
{

    int a[100001];
    int i,k,t,flag;
    long long int min;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            cin >> a[i];
        }


        for(k=1;k<n;k++)
        {
            min=0;
            for(i=1;i<=k;i++)
            {
                flag=bfs(i,k,0);
                min=min+flag*a[i];
            }
            /*t=log(n-k)/log(2);
            min=min+a[k];
            a[k+pp(2,t)]+=a[k];
            a[k]=0;*/

            //min=min+a[i];
            cout << min <<endl;
        }
    }

    return 0;
}
