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>
using namespace std;
inline int add(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  if (_a + _b >= 1000000007) {
    return _a + _b - 1000000007;
  }
  return _a + _b;
}
inline int mul(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  return ((long long int)((long long int)_a * (long long int)_b)) % 1000000007;
}
const int N = 1000000;
int n, ara[N + 10];
double sol[N + 10];
long long int cross(pair<int, long long int> p0, pair<int, long long int> p1,
                    pair<int, long long int> p2) {
  int x1 = p1.first - p0.first;
  long long int y1 = p1.second - p0.second;
  int x2 = p2.first - p0.first;
  long long int y2 = p2.second - p0.second;
  return y2 * (long long int)x1 - (long long int)x2 * y1;
}
void solve() {
  int i, j, sz;
  long long int sum = 0;
  vector<pair<int, long long int> > hull;
  hull.push_back(make_pair(0, 0));
  for (i = 0; i < n; ++i) {
    sum += ara[i];
    while ((sz = hull.size()) > 1 &&
           cross(hull[sz - 2], hull[sz - 1], make_pair(i + 1, sum)) <= 0) {
      hull.pop_back();
    }
    hull.push_back(make_pair(i + 1, sum));
  }
  for (i = 0; i < hull.size(); ++i) {
    int th_x = i == 0 ? 0 : hull[i - 1].first;
    long long int th_y = i == 0 ? 0 : hull[i - 1].second;
    for (j = hull[i].first - 1; j >= (i == 0 ? 0 : hull[i - 1].first); --j) {
      sol[j] = (double)(hull[i].second - th_y) / (hull[i].first - th_x);
    }
  }
  for (i = 0; i < n; i++) {
    printf("%.10f\n", sol[i]);
  }
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &ara[i]);
  }
  solve();
}