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;
struct point {
  char ch;
  int num, id;
} pnt[100050];
struct ID {
  int num;
  int id;
} pt[100050];
bool cmp1(ID a1, ID b1) {
  if (a1.num != b1.num) return a1.num < b1.num;
  return a1.id < b1.id;
}
int hs[100050], pr[100050], id[100050], cnt, cnt1, ans, n, m, k;
int bs(int tmp, int tmpid) {
  int fr, ra, md;
  fr = 0, ra = k - 1;
  while (fr <= ra) {
    md = (fr + ra) / 2;
    if (pt[md].id == tmpid && pt[md].num == tmp) return md;
    if (pt[md].num > tmp)
      ra = md - 1;
    else if (pt[md].num == tmp && pt[md].id > tmpid)
      ra = md - 1;
    else
      fr = md + 1;
  }
}
int find_set(int id) {
  int tmp;
  tmp = pr[id];
  while (tmp < id || (tmp - id) % m != 0) tmp += n;
  ans += (tmp - id) / m;
  if (id != pr[id]) pr[id] = find_set(pr[id]);
  return pr[id];
}
int main() {
  int i, tmp, tmpid, fr, ra, md, besti;
  cnt = 1, ans = 0;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i <= n; i++) pr[i] = i, hs[i] = 0;
  for (i = 0; i < k; i++) {
    getchar();
    scanf("%c", &pnt[i].ch);
    if (pnt[i].ch == '-') {
      scanf("%d", &pnt[i].num);
      pnt[i].id = i;
      pt[i].num = pnt[i].num;
      pt[i].id = i;
    } else {
      scanf("%d%d", &pnt[i].num, &pnt[i].id);
      pt[i].num = pnt[i].num;
      pt[i].id = i;
    }
  }
  sort(pt, pt + k, cmp1);
  for (i = 0; i < k; i++) {
    if (pnt[i].ch == '-') {
      tmp = pnt[i].num;
      tmpid = i;
      besti = bs(tmp, tmpid);
      besti -= 1;
      tmp = hs[besti];
      hs[besti] = -1;
      pr[tmp] = tmp, fr = (tmp - m + n) % n;
      while (pr[fr] != fr) {
        pr[fr] = tmp;
        fr = (fr - m + n) % n;
      }
    } else {
      tmp = pnt[i].num;
      tmpid = i;
      besti = bs(tmp, tmpid);
      fr = find_set(pnt[i].id);
      hs[besti] = fr;
      pr[fr] = (fr + m) % n;
    }
  }
  printf("%d\n", ans);
  return 0;
}
