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;
const long long int INF = 9223372036854775807;
const long long int mod = 998244353;
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
struct point {
  long long int x;
  long long int y;
};
long long int cross(point a, point b) { return a.x * b.y - b.x * a.y; }
bool ccw1(point p1, point p2) {
  if (cross(p1, p2) > 0) {
    return true;
  } else {
    return false;
  }
}
bool comp(point p1, point p2) {
  if (p1.x < p2.x) {
    return true;
  } else {
    if (p1.x > p2.x) {
      return false;
    } else {
      if (p1.y <= p2.y) {
        return true;
      }
      return false;
    }
  }
}
long long ccw(point p1, point p2, point p3) {
  long long int val =
      (p2.y - p1.y) * (p3.x - p2.x) - (p2.x - p1.x) * (p3.y - p2.y);
  return val;
}
void ccw_sort(vector<point> &v) {
  vector<point> v1, v2;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].y > 0) {
      v1.push_back(v[i]);
    } else {
      v2.push_back(v[i]);
    }
  }
  sort(v1.begin(), v1.end(), ccw1);
  sort(v2.begin(), v2.end(), ccw1);
  for (int i = 0; i < v2.size(); i++) {
    v1.push_back(v2[i]);
  }
  v = v1;
}
vector<point> lower_convex_hull(vector<point> pts) {
  sort(pts.begin(), pts.end(), comp);
  vector<point> H;
  for (int step = 0; step < 1; step++) {
    auto start = H.size();
    for (point P : pts) {
      while (H.size() >= start + 2 &&
             ccw(H[H.size() - 2], H[H.size() - 1], P) <= 0)
        H.pop_back();
      H.push_back(P);
    }
  }
  return H;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  vector<point> g;
  long long int sum = 0;
  point cur;
  cur.x = cur.y = 0;
  g.push_back(cur);
  for (int i = 0; i < n; i++) {
    sum += a[i];
    point pt;
    pt.x = sum;
    pt.y = i + 1;
    g.push_back(pt);
  }
  auto ans = lower_convex_hull(g);
  vector<double> v;
  for (int j = 1; j < ans.size(); j++) {
    for (long long int yy = ans[j - 1].y + 1; yy <= ans[j].y; yy++) {
      v.push_back((double)(ans[j].x - ans[j - 1].x) /
                  (double)(ans[j].y - ans[j - 1].y));
    }
  }
  for (int i = 0; i < v.size(); i++) {
    printf("%.12f\n", v[i]);
  }
}