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 <bits/stdc++.h>
using namespace std;
long n, m;
long a[200001], w[200001], d[200001];
long t[100];
int main() {
  long i, j, k;
  t[0] = 1;
  i = 1;
  k = 0;
  while (1) {
    t[i] = t[i - 1] * 2;
    if (t[i] > 120000) break;
    i++;
    k++;
  }
  while (scanf("%d", &n) != -1) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    memset(w, 0, sizeof(w));
    memset(d, 0, sizeof(d));
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= k; j++) {
        if (i + t[j] > n) {
          w[i + t[j - 1]] += a[i] + w[i];
          break;
        }
      }
    }
    d[1] = a[1];
    for (i = 2; i <= n - 1; i++) {
      d[i] = d[i - 1] + a[i] + w[i];
    }
    for (i = 1; i < n; i++) printf("%d\n", d[i]);
  }
}
