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;
int main() {
  char N[10];
  cin.getline(N, 10);
  int ch = 0;
  int p = 0;
  while (N[p++] != '\0') ch++;
  int n = 0;
  int ch_temp = ch - 1;
  for (int i = 0; i < ch; i++) n = n + (int)(N[i] - '0') * pow(10., ch_temp--);
  int* mass = new int[n];
  int* mass1 = new int[n];
  for (int i = 0; i < n; i++) cin >> mass[i];
  for (int k = 0; k < n - 1; k++) {
    int sc = 0;
    for (int i = 0; i < n; i++) mass1[i] = mass[i];
    for (int i = 0; i <= k; i++) {
      int s = 0;
      for (int t = 0; t < n; t++) {
        if (i + pow(2., t) > k && i + pow(2., t) < n) {
          s = i + pow(2., t);
          break;
        }
      }
      if (s == 0) {
        for (int t = 0; t < n; t++) {
          if (i + 1 + pow(2., t) > k && i + 1 + pow(2., t) < n) {
            s = i + 1;
            break;
          }
        }
      }
      if (s == 0) {
        for (int j = i + 2; j <= k; j++) {
          j++;
          s = 0;
          for (int t = 0; t < n; t++) {
            if (j + pow(2., t) > k && j + pow(2., t) < n) {
              s = j;
              break;
            }
          }
          if (s != 0) break;
        }
      }
      while (mass1[i] != 0) {
        mass1[i]--;
        mass1[s]++;
        sc++;
      }
    }
    cout << sc << endl;
  }
  delete[] mass;
  delete[] mass1;
  getchar();
  getchar();
  return 0;
}
