Problem: There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.

You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.

What is the lexicographically smallest sequence of volumes of water that you can achieve?

As a reminder:

A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.

Because of large input, reading input as doubles is not recommended.

Output

Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.

Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.

Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.

Examples

Input


4
7 5 5 7


Output


5.666666667
5.666666667
5.666666667
7.000000000


Input


5
7 8 8 10 12


Output


7.000000000
8.000000000
8.000000000
10.000000000
12.000000000


Input


10
3 9 5 5 1 7 5 3 8 7


Output


3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000

Note

In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].

In the second sample, you can't get any lexicographically smaller sequence.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long n, kl, m, k, i, j, h, a[1000001], b[1000001];
long double pr[1000001];
long double g[1000001];
string s;
vector<pair<pair<long double, long double>, long double> > v;
long double get_cross_x(pair<long double, long double> x,
                        pair<long double, long double> y) {
  return (x.second - y.second) / (y.first - x.first);
}
void add(long double k, long double x) {
  while (v.size() > 0) {
    long double p =
        get_cross_x(make_pair((long double)k, (long double)x), v.back().first);
    if (p - 1e-12 > v.back().second)
      v.pop_back();
    else {
      v.push_back(make_pair(make_pair(k, x), p));
      break;
    }
  }
  if (v.size() == 0) {
    v.push_back(make_pair(make_pair(k, x), 1000000000000000000));
    return;
  }
}
pair<pair<long double, long long>, int> get(long double c) {
  int l = 0, r = v.size() - 1;
  while (l < r) {
    int w = (l + r + 1) / 2;
    if (v[w].second >= c)
      l = w;
    else
      r = w - 1;
  }
  return make_pair(make_pair(((long double)c * (long double)v[l].first.first) +
                                 v[l].first.second,
                             -(long long)v[l].first.first),
                   l);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pr[i] = pr[i - 1] + a[i];
  }
  for (int i = n; i > 0; i--) {
    add((long double)-i, (long double)pr[i]);
  }
  long long r = v.size() - 1;
  long double lr, rr, w;
  for (int i = 1; i <= n;) {
    while ((long long)-v[r].first.first < i) r--;
    while (r > 0) {
      long double p = v[r].second;
      if (v[r].first.first * p + v[r].first.second >=
          -((long double)i - 1) * p + (long double)pr[i - 1] - 1e-12)
        r--;
      else
        break;
    }
    long long j = (long long)-v[r].first.first;
    for (int u = i; u <= j; u++)
      g[u] = (long double)(pr[j] - pr[i - 1]) / (long double)(j - i + 1);
    i = j + 1;
  }
  cout.precision(10);
  for (int i = 1; i <= n; i++) cout << fixed << g[i] << "\n";
  return 0;
}