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;
FILE *in;
FILE *out;
int n;
int a[131072];
pair<int, int> tree[262144];
void update(int idx, pair<int, int> val) {
  for (idx = idx + (262144 >> 1); idx; idx >>= 1)
    tree[idx] = min(tree[idx], val);
}
pair<int, int> query(int idx1, int idx2) {
  idx1 += (262144 >> 1);
  idx2 += (262144 >> 1);
  if (idx1 == idx2) return tree[idx1];
  pair<int, int> ret = make_pair(999666333, 0);
  ret = min(ret, tree[idx1]);
  int flag1 = !(idx1 & 1);
  idx1 >>= 1;
  ret = min(ret, tree[idx2]);
  int flag2 = (idx2 & 1);
  idx2 >>= 1;
  while (idx1 != idx2) {
    if (flag1) ret = min(ret, tree[(idx1 << 1) + 1]);
    if (flag2) ret = min(ret, tree[(idx2 << 1) + 0]);
    flag1 = !(idx1 & 1);
    idx1 >>= 1;
    flag2 = (idx2 & 1);
    idx2 >>= 1;
  }
  return ret;
}
int popcnt(int n) {
  int ret = 0;
  while (n) n &= (n - 1), ret++;
  return ret;
}
string toString(long long n) {
  if (n == 0) return "0";
  string ret;
  while (n) {
    ret.push_back(n % 10 + 48);
    n /= 10;
  }
  reverse(ret.begin(), ret.end());
  return ret;
}
queue<pair<int, int> > q[131072];
int main(void) {
  in = stdin;
  out = stdout;
  fscanf(in, "%d", &n);
  for (int i = 1; i <= n; i++) fscanf(in, "%d", &a[i]);
  for (int i = 0; i < 262144; i++) tree[i] = make_pair(999666333, 0);
  for (int i = 1; i <= n; i++) update(i, make_pair(popcnt(i), -i));
  long long ans = 0;
  for (int k = 1; k < n; k++) {
    pair<int, int> minn = query(1, n - k);
    ans += minn.first * a[k];
    q[k - minn.second].push(make_pair(minn.first, k));
    while (!q[k].empty()) {
      pair<int, int> cur = q[k].front();
      q[k].pop();
      ans -= cur.first * a[cur.second];
      minn = query(k - cur.second + 1, n - cur.second);
      ans += minn.first * a[cur.second];
      q[cur.second - minn.second].push(make_pair(minn.first, cur.second));
    }
    fprintf(out, "%s\n", toString(ans).c_str());
  }
  return 0;
}
