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;
const double pi = acos(-1.0);
const int inf = (int)1e9;
const long long int infll = (long long int)1e18;
const double inf1 = 1e-9;
const int ss = (int)1e6 + 3;
const int base = inf;
bool pred(const pair<int, int>& i, const pair<int, int>& j) {
  if (i.first == j.first)
    return i.second > j.second;
  else
    return i.first < j.first;
}
bool pred2(const int& i, const int& j) { return i > j; }
vector<vector<int> > f;
int n;
int k;
vector<int> comp;
vector<bool> used;
long long int mx = 0;
vector<vector<bool> > e;
void dfs(int v) {
  used[v] = true;
  comp.push_back(v);
  for (int i = 0; i < f[v].size(); ++i) {
    int to = f[v][i];
    if (!used[to]) dfs(to);
  }
}
void find_comps() {
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      comp.clear();
      dfs(i);
      if (comp.size() > mx) {
        bool t = true;
        for (int j = 0; j < comp.size() - 1; ++j) {
          for (int l = j + 1; l < comp.size(); ++l) {
            if (e[comp[j]][comp[l]]) {
              t = false;
              break;
            }
          }
          if (!t) break;
        }
        if (t) mx = comp.size();
      }
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<long long int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  int z = 1;
  vector<int> mas;
  while (z < n) {
    mas.push_back(z);
    z *= 2;
  }
  mas.push_back(z);
  vector<int> b(n);
  z = 0;
  long long int res = 0;
  for (int i = 0; i < n - 1; ++i) {
    z = -1;
    do {
      ++z;
    } while (i + mas[z] < n);
    --z;
    long long int t = a[i];
    a[i + mas[z]] += t;
    res += t;
    printf("%d\n", res);
  }
  return 0;
}
