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;
template <class T>
inline void rd(T &x) {
  char ch;
  x = 0;
  bool fl = false;
  while (!isdigit(ch = getchar())) (ch == '-') && (fl = true);
  for (x = (ch ^ '0'); isdigit(ch = getchar()); x = x * 10 + (ch ^ '0'))
    ;
  (fl == true) && (x = -x);
}
template <class T>
inline void output(T x) {
  if (x / 10) output(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void ot(T x) {
  if (x < 0) putchar('-'), x = -x;
  output(x);
  putchar(' ');
}
template <class T>
inline void prt(T a[], int st, int nd) {
  for (register int i = st; i <= nd; ++i) ot(a[i]);
  putchar('\n');
}
namespace Modulo {
const int mod = 998244353;
inline int ad(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int sub(int x, int y) { return ad(x, mod - y); }
inline int mul(int x, int y) { return (long long)x * y % mod; }
inline void inc(int &x, int y) { x = ad(x, y); }
inline void inc2(int &x, int y) { x = mul(x, y); }
inline int qm(int x, int y = mod - 2) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = mul(x, ret);
    x = mul(x, x);
    y >>= 1;
  }
  return ret;
}
template <class... Args>
inline int ad(const int a, const int b, const Args &...args) {
  return ad(ad(a, b), args...);
}
template <class... Args>
inline int mul(const int a, const int b, const Args &...args) {
  return mul(mul(a, b), args...);
}
}  // namespace Modulo
namespace Miracle {
const int N = 1e6 + 5;
struct po {
  long double x, y;
  po() {}
  po(long double xx, long double yy) {
    x = xx;
    y = yy;
  }
  po friend operator-(const po &a, const po &b) {
    return po(a.x - b.x, a.y - b.y);
  }
} p[N];
int a[N];
long long s[N];
long double ans[N];
int n;
int sta[N], top;
long double cha(po A, po B) { return A.x * B.y - A.y * B.x; }
int main() {
  rd(n);
  for (int i = 1; i <= n; ++i) {
    rd(a[i]);
    s[i] = s[i - 1] + a[i];
    p[i].x = i, p[i].y = s[i];
  }
  p[0].x = 0;
  p[0].y = 0;
  sta[++top] = 0;
  for (int i = 1; i <= n; ++i) {
    while (top > 1 &&
           cha(p[i] - p[sta[top - 1]], p[sta[top]] - p[sta[top - 1]]) > 0)
      --top;
    sta[++top] = i;
  }
  for (int i = 1; i < top; ++i) {
    long double sum = s[sta[i + 1]] - s[sta[i]];
    long double now = sum / (sta[i + 1] - sta[i]);
    for (int j = sta[i] + 1; j <= sta[i + 1]; ++j) {
      ans[j] = now;
    }
  }
  for (int i = 1; i <= n; ++i) {
    printf("%.9Lf\n", ans[i]);
  }
  return 0;
}
}  // namespace Miracle
signed main() {
  Miracle::main();
  return 0;
}