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;
string to_string(string c) { return string(c); }
string to_string(char ch) { return string(1, ch); }
template <size_t N>
string to_string(bitset<N> B) {
  string second = "";
  for (long long i = 0; i < (long long)N; i++) second += (char)'0' + B[i];
  return second;
}
template <typename T, typename U>
string to_string(pair<T, U> P) {
  return "{" + to_string(P.first) + "," + to_string(P.second) + "}";
}
template <typename T>
string to_string(T A) {
  string second = "[";
  bool first = 0;
  for (auto x : A) {
    if (first) second += " ";
    first = 1;
    second += to_string(x);
  }
  return second + "]";
}
template <typename T, size_t N>
string to_string(T (&A)[N]) {
  string second = "[";
  bool first = 0;
  for (long long i = 0; i < N; i++) {
    if (first) second += " ";
    first = 1;
    second += to_string(A[i]);
  }
  return second + "]";
}
void _deb() {
  cerr << "]"
       << "\n";
}
template <typename T, typename... U>
void _deb(const T& t, const U&... u) {
  cerr << to_string(t);
  if (sizeof...(u)) cout << ", ";
  _deb(u...);
}
long long ceil(long long a, long long b) {
  return a / b + (a % b && (a ^ b) > 0);
}
long long floor(long long a, long long b) {
  return a / b - (a % b && (a ^ b) < 0);
}
long long SBITS(long long n) { return __builtin_popcountll(n); }
long long BITS(long long n) { return 63 - __builtin_clzll(n); }
const long long MX = 5e5 + 5;
const long long OO = 1e18;
const long long MOD = 1e9 + 7;
long long DX[8] = {1, 0, -1, 0, 1, -1, 1, -1};
long long DY[8] = {0, 1, 0, -1, 1, -1, -1, 1};
void solve() {
  long long n, ans = 0;
  cin >> n;
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i < n; i++) {
    ans += a[i];
    long long maxN = log2(n - i);
    a[i + (1 << maxN)] += a[i];
    cout << ans << "\n";
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
