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 JAVA):

Solution:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class D_CF {

    public static void main(String[] args) {
        Long X = (long)(1e9+1);
        FastScanner58 fs = new FastScanner58();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        // for (int tc = 0; tc < t; tc++) {
        for (int tc = 0; tc < t; tc++) {
           int n = fs.ni();
           long[] a = fs.longArray(n);
           long s = 0;
           long[] ps = new long[n];
           for (int i = 0; i < n; i++) {
               s += a[i];
               ps[i] = s;
           }
           Stack<Integer> pos = new Stack();
           pos.add(-1);
           double[] res = new double[n];
           for (int i = 0; i < n; i++) {
               while (pos.size()>1) {
                   int p = pos.pop();
                   int op = pos.peek();
                   int np = i;
                   double nw = ps[np];
                   if (op!=-1) nw -= ps[op];
                   nw /= (double)(np-op);
                   double cw = ps[p];
                   if (op!=-1) cw -= ps[op];
                   cw /= (double)(p-op);
                   if (nw<cw) {
                       continue;
                   } else {
                       pos.add(p);
                       pos.add(np);
                       break;
                   }
               }
               if (pos.size()==1) {
                   pos.add(i);
               }
           }
           int p = n;
           for (int i = n-1; i > -1; i--) {
               if (pos.peek()==i) {
                   p = pos.pop();
               }
               double d = ps[p];
               if (pos.peek()>-1) d-=ps[pos.peek()];
               d/=p-pos.peek();
               res[i] = d;
           }
           for (int i = 0; i < n; i++) {
               pw.println(res[i] + " ");
           }
        }
        pw.close();
    }
    public static long dis(int i, int j, long[] x, long[] y) {
        return (long)Math.pow(Math.abs(x[i]-x[j]),2) + (long)Math.pow(Math.abs(y[i]-y[j]),2);
    }
    public static long gcd(long n1, long n2) {
        if (n2 == 0) {
            return n1;
        }
        return gcd(n2, n1 % n2);
    }
}
class UnionFind18 {

    int[] id;

    public UnionFind18(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}

class FastScanner58 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner58() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

class UnionFind17 {

    int[] id;

    public UnionFind17(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}