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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int MAXN = 100100;
long long res[MAXN];
int n, a[MAXN];
vector<int> v[MAXN];
void add(long long value, int l, int r) {
  if (l > r) return;
  if (l < MAXN) res[l] += value;
  if (r < MAXN) res[r] -= value;
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = 0; i < (int)(1 << 20); i++) {
    int cur = 0;
    for (int j = 0; j < (int)(20); j++)
      if ((i & (1 << j)) != 0) cur++;
    v[cur].push_back(i);
  }
  for (int i = 0; i < (int)(n - 1); i++) {
    int l = i;
    for (int r = 1; r < 20; r++) {
      int pos = (int)(upper_bound((v[r]).begin(), (v[r]).end(), n - i - 1) -
                      v[r].begin()) -
                1;
      if (pos >= 0) {
        pos = v[r][pos];
        add(a[i] * r, l, i + pos);
        l = i + pos;
      }
    }
  }
  long long cur = 0;
  for (int i = 0; i < (int)(n - 1); i++) {
    cur += res[i];
    printf("%I64d ", cur);
  }
  return 0;
}
