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;
using pii = pair<int, int>;
using vpii = vector<pii>;
using vi = vector<int>;
using vvi = vector<vi>;
using ll = long long;
using vll = vector<long long>;
template <class T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
istream &operator>>(istream &, vector<T> &);
template <class T>
ostream &operator<<(ostream &, const vector<T> &);
template <class T, class U>
istream &operator>>(istream &, pair<T, U> &);
template <class T, class U>
ostream &operator<<(ostream &, const pair<T, U> &);
template <class T>
struct Inf {
  static constexpr T inf() {
    return std::numeric_limits<T>::has_infinity
               ? std::numeric_limits<T>::infinity()
               : std::numeric_limits<T>::max();
  }
};
template <>
struct Inf<int> {
  static constexpr int inf() { return 0x3f3f3f3f; }
};
template <>
struct Inf<long long> {
  static constexpr long long inf() { return 0x3f3f3f3f3f3f3f3fLL; }
};
constexpr int INF = Inf<int>::inf();
constexpr ll BINF = Inf<ll>::inf();
int solve() {
  int n;
  cin >> n;
  vll a(n);
  cin >> a;
  ll ans = 0;
  for (int i = 0; i + 1 < n; i++) {
    for (int t = 20; t >= 0; t--) {
      if (i + (1 << t) >= n) continue;
      ans += a[i];
      a[i + (1 << t)] += a[i];
      break;
    }
    cout << ans << endl;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto it = v.begin(); it != v.end(); ++it) is >> *it;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto it = v.begin(); it != v.end();)
    os << *it, os << " \n"[++it == v.end()];
  return os;
}
template <class T, class U>
istream &operator>>(istream &is, pair<T, U> &p) {
  return is >> p.first >> p.second;
}
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << p.first << ' ' << p.second;
}
