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 int N = 100010;
long long val[N], sum[N], dp[N];
int n;
bool used[N];
queue<int> q;
vector<int> G[N];
vector<long long> ans;
void bfs(int st) {
  used[st] = true;
  q.push(st);
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; v - (1 << i) > 0; i++)
      if (used[v - (1 << i)] == false) {
        used[v - (1 << i)] = true;
        q.push(v - (1 << i));
        G[v].push_back(v - (1 << i));
      }
  }
}
void dfs(int v) {
  used[v] = true;
  sum[v] = val[v];
  dp[v] = 0;
  for (int i = 0; i < (int)(G[v]).size(); ++i) {
    int to = G[v][i];
    if (used[to] == true) continue;
    dfs(to);
    sum[v] += sum[to];
    dp[v] += dp[to] + sum[to];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &val[i]);
  for (int i = 1; i <= n; ++i) used[i] = false;
  bfs(n);
  for (int i = 1; i <= n; ++i) used[i] = false;
  dfs(n);
  ans.push_back(dp[n]);
  long long k = dp[n];
  for (int i = n - 1; i > 1; --i) {
    k -= sum[i];
    ans.push_back(k);
  }
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < (int)(ans).size(); ++i) printf("%I64d ", ans[i]);
  return 0;
}
