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;
static const double EPS = 1e-10;
template <typename T, typename S>
vector<T>& operator<<(vector<T>& a, S b) {
  a.push_back(b);
  return a;
}
template <typename T>
void operator>>(vector<T>& a, int b) {
  while (b--)
    if (!a.empty()) a.pop_back();
}
bool isprime(int n) {
  if (n < 2) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long b_pow(long long x, long long n) {
  return n ? b_pow(x * x, n / 2) * (n % 2 ? x : 1) : 1ll;
}
string itos(int n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
int a[100001];
int t[100001];
int n;
long long solve(int p) {
  for (int i = 0; i < n; i++) a[i] = t[i];
  long long ans = 0;
  for (int i = 0; i <= p; i++) {
    int u = -1, d = i + 1;
    for (int j = 0; i + (1 << j) < n; j++) {
      if (i + (1 << j) > p) {
        u = (i + (1 << j));
      } else {
        d = i + (1 << j);
      }
    }
    if (u == -1) {
      ans += a[i];
      a[d] += a[i];
      a[i] = 0;
    } else {
      ans += a[i];
      a[u] += a[i];
      a[i] = 0;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n - 1; i++) {
    cout << solve(i) << endl;
  }
}
