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 INFL = 1LL << 60;
const long long INF = 1 << 30;
long long MOD = 1000000007;
vector<long long> dy = {0, 1, 0, -1, 1, 1, -1, -1, 0};
vector<long long> dx = {1, 0, -1, 0, 1, -1, 1, -1, 0};
void pres(double A) { printf("%.9lf\n", A); }
void BinarySay(long long x, long long y = 60) {
  for (long long i = 0; i < (y); i++) cout << (x >> (y - 1 - i) & 1);
  cout << endl;
}
long long cnt_bit(long long x) { return __builtin_popcountll(x); }
long long pow_long(long long x, long long k) {
  long long res = 1;
  while (k > 0) {
    if (k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
long long pow_mod(long long x, long long k) {
  long long res = 1;
  while (k > 0) {
    if (k % 2) {
      res *= x;
      res %= MOD;
    }
    x *= x;
    x %= MOD;
    k /= 2;
  }
  return res;
}
long long inverse(long long x) { return pow_mod(x, MOD - 2); };
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long x, long long y) {
  long long res = x / gcd(x, y);
  res *= y;
  return res;
};
const int MAXcomb = 2000010;
long long fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAXcomb; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}
long long comb(int n, int k) {
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * finv[k] % MOD * finv[n - k] % MOD;
}
const int MAXkai = 200010;
long long kai_memo[MAXkai];
long long kai(long long N) {
  if (kai_memo[N] != 0) return kai_memo[N];
  if (N <= 1) return 1;
  return kai_memo[N] = N * kai(N - 1) % MOD;
}
void solve() {
  long long N;
  cin >> N;
  long long a;
  vector<pair<long long, long long>> A(N);
  for (long long i = 0; i < (N); i++) {
    cin >> A[i].first;
    A[i].second = 1;
  }
  vector<long long> left(N + 2), right(N + 2);
  for (long long i = 0; i < (N + 2); i++) left[i] = i - 1;
  for (long long i = 0; i < (N + 2); i++) right[i] = i + 1;
  long long now = 1;
  while (now <= N && right[now] <= N) {
    if (A[now - 1].first * A[right[now] - 1].second >=
        A[right[now] - 1].first * A[now - 1].second) {
      A[now - 1].first += A[right[now] - 1].first;
      A[now - 1].second += A[right[now] - 1].second;
      right[now] = right[right[now]];
      left[right[now]] = now;
      while (now >= 1 && left[now] >= 1) {
        if (A[left[now] - 1].first * A[now - 1].second >=
            A[now - 1].first * A[left[now] - 1].second) {
          A[now - 1].first += A[left[now] - 1].first;
          A[now - 1].second += A[left[now] - 1].second;
          left[now] = left[left[now]];
          right[left[now]] = now;
        } else {
          break;
        }
      }
    } else {
      now = right[now];
    }
  }
  vector<pair<long long, long long>> ans;
  while (now >= 1) {
    long long last = now, ne = left[now];
    while (now > N) now = left[now];
    for (long long _ = 0; _ < (A[now - 1].second); _++) {
      ans.push_back({A[now - 1].first, A[now - 1].second});
    }
    now = left[now];
  }
  for (long long i = (ans.size()) - 1; i >= 0; i--) {
    double x = 1.0 * ans[i].first / ans[i].second;
    pres(x);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}