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 = 3.141592653589793;
const int INF = 2000000000;
const int mod = 1000000007;
const int N = 100000000;
const int base = 10;
vector<int> read(string s) {
  vector<int> a;
  for (int i = (int)s.length(); i > 0; i -= 1)
    if (i < 1)
      a.push_back(atoi(s.substr(0, i).c_str()));
    else
      a.push_back(atoi(s.substr(i - 1, 1).c_str()));
  return a;
}
vector<int> add(vector<int> a, vector<int> b) {
  int carry = 0;
  for (size_t i = 0; i < max(a.size(), b.size()) || carry; ++i) {
    if (i == a.size()) a.push_back(0);
    a[i] += carry + (i < b.size() ? b[i] : 0);
    carry = a[i] >= base;
    if (carry) a[i] -= base;
  }
  return a;
}
vector<int> mul(vector<int> a, int b) {
  int carry = 0;
  for (size_t i = 0; i < a.size() || carry; ++i) {
    if (i == a.size()) a.push_back(0);
    long long cur = carry + a[i] * 1ll * b;
    a[i] = int(cur % base);
    carry = int(cur / base);
  }
  while (a.size() > 1 && a.back() == 0) a.pop_back();
  return a;
}
int main() {
  long long a[10005], b[10005];
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = 0;
  }
  b[0] = 0;
  for (int(i) = (1); (i) < (n); (i)++) {
    b[i] += a[i] + b[i - 1];
    int j = 1;
    while (j + i <= n) {
      j = j << 1;
    }
    j = j >> 1;
    a[i + j] += a[i];
  }
  for (int(j) = (1); (j) < (n); (j)++) cout << b[j] << endl;
  return 0;
}
