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>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int pct(int x) { return __builtin_popcount(x); }
int bits(int x) { return 31 - __builtin_clz(x); }
int fstTrue(function<bool(int)> first, int lo, int hi) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
const long long MOD = 1e9 + 7;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long modInv(long long a) { return binpow(a, MOD - 2); }
bool sortcol(const vector<long long>& v1, const vector<long long>& v2) {
  return v1[1] < v2[1];
}
bool sortpair(const pair<int, int>& p1, const pair<int, int>& p2) {
  return p1.first < p2.first;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<long long unsigned> distribution(0, 10);
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int first = 0;
  cerr << '{';
  for (auto& i : x) cerr << (first++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long double PI = 2.0 * acos(0.0);
const long double EPS = 1e-9;
struct PT {
  long double x, y;
  long double length() { return sqrt(x * x + y * y); }
  int normalize() {
    long double l = length();
    if (fabs(l) < EPS) return -1;
    x /= l;
    y /= l;
    return 0;
  }
  PT operator-(PT a) {
    PT r;
    r.x = x - a.x;
    r.y = y - a.y;
    return r;
  }
  PT operator+(PT a) {
    PT r;
    r.x = x + a.x;
    r.y = y + a.y;
    return r;
  }
  PT operator*(long double sc) {
    PT r;
    r.x = x * sc;
    r.y = y * sc;
    return r;
  }
};
bool operator<(const PT& a, const PT& b) {
  if (fabs(a.x - b.x) < EPS) return (a.y < b.y);
  return (a.x < b.x);
}
long double dist(PT& a, PT& b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
long double dot(PT& a, PT& b) { return (a.x * b.x + a.y * b.y); }
long double cross(PT& a, PT& b) { return (a.x * b.y - a.y * b.x); }
bool cw(PT a, PT b, PT c) {
  return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) < 0;
}
bool ccw(PT a, PT b, PT c) {
  return a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y) >= 0;
}
int intersection(PT p1, PT p2, PT p3, PT p4, PT& r) {
  long double d = (p4.y - p3.y) * (p2.x - p1.x) - (p4.x - p3.x) * (p2.y - p1.y);
  if (fabs(d) < EPS) return -1;
  long double ua, upper_bound;
  ua = (p4.x - p3.x) * (p1.y - p3.y) - (p4.y - p3.y) * (p1.x - p3.x);
  ua /= d;
  r = p1 + (p2 - p1) * ua;
  return 0;
}
const int mxN = 1e6 + 5;
int n;
vector<PT> arr;
vector<PT> graham_scan(vector<PT> p) {
  sort(begin(p), end(p));
  vector<PT> chull(p.size() + 1);
  int idx = 0;
  for (int i = (0); i < (p.size()); i++) {
    while (idx >= 2 && ccw(chull[idx - 2], chull[idx - 1], p[i])) idx--;
    chull[idx++] = p[i];
  }
  int half = idx;
  for (int i = (p.size() - 1) - 1; i >= (0); i--) {
    while (idx > half && ccw(chull[idx - 2], chull[idx - 1], p[i])) idx--;
    chull[idx++] = p[i];
  }
  idx--;
  chull.resize(idx);
  return chull;
}
void solve() {
  cin >> n;
  arr.push_back({0, 0});
  long double curr = 0;
  for (int i = (1); i < (n + 1); i++) {
    arr.push_back({(long double)i, 0});
    int y;
    cin >> y;
    curr += y;
    arr[i].y = curr;
  }
  vector<PT> chull = graham_scan(arr);
  chull.push_back({0, 0});
  reverse(begin(chull), end(chull));
  while (chull[chull.size() - 1].y < arr[n].y) chull.pop_back();
  for (int i = (0); i < (chull.size() - 1); i++)
    for (int j = (chull[i].x); j < (chull[i + 1].x); j++)
      cout << setprecision(18)
           << (chull[i + 1].y - chull[i].y) / (chull[i + 1].x - chull[i].x)
           << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) solve();
}