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;
bool SR(int &x) { return scanf("%d", &x) == 1; }
bool SR(long long &x) { return scanf("%lld", &x) == 1; }
bool SR(double &x) { return scanf("%lf", &x) == 1; }
bool SR(char *s) { return scanf("%s", s) == 1; }
bool RI() { return true; }
template <typename I, typename... T>
bool RI(I &x, T &...tail) {
  return SR(x) && RI(tail...);
}
void SP(const int x) { printf("%d", x); }
void SP(const long long x) { printf("%lld", x); }
void SP(const double x) { printf("%.16lf", x); }
void SP(const char *s) { printf("%s", s); }
void PL() { puts(""); }
template <typename I, typename... T>
void PL(const I x, const T... tail) {
  SP(x);
  if (sizeof...(tail)) putchar(' ');
  PL(tail...);
}
const int maxn = 1e5 + 5;
int n, a[maxn];
void read() {
  RI(n);
  for (int i = (1); i <= int(n); i++) RI(a[i]);
}
inline int count(const int &x) { return __builtin_popcount(x); }
vector<int> mov[maxn];
void build() {
  for (int i = (1); i <= int(n - 1); i++) {
    vector<int> &v = mov[i];
    for (int j = n - i; j; j -= j & -j) v.push_back(j);
    reverse((v).begin(), (v).end());
  }
}
inline int det(const pair<int, int> &x) {
  return mov[x.first][x.second] + x.first;
}
inline long long cost(const pair<int, int> &x) {
  return 1LL * a[x.first] * count(mov[x.first][x.second]);
}
long long s[maxn];
void sol() {
  for (int i = (1); i <= int(n - 1); i++) {
    s[i] += 1LL * a[i] * count(mov[i][0]);
    for (int j = 0; j < int(((int)(mov[i]).size()) - 1); j++)
      s[i + mov[i][j]] +=
          1LL * a[i] * (count(mov[i][j + 1]) - count(mov[i][j]));
  }
  for (int i = 0; i < int(n); i++) s[i + 1] += s[i];
  for (int i = (1); i <= int(n - 1); i++) PL(s[i]);
}
int main() {
  read();
  build();
  sol();
  return 0;
}
