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:
import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.nio.charset.StandardCharsets;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.InputMismatchException;

public class C1299 {
	static class Solver {
		
		int N;
		long a[];
		ArrayDeque<Block> res = new ArrayDeque<>();
		
		Block first, second;
		void collapse() {
			while(res.size() > 1) {
				first = res.pollFirst();
				second = res.pollFirst();
				if(first.shouldMerge(second)) {
					first.merge(second);
					res.addFirst(first);
				} else {
					res.addFirst(second);
					res.addFirst(first);
					break;
				}
			}
		}
		
		void solve(int testNumber, FastScanner s, PrintWriter out) {
			N = s.nextInt(); a = s.nextLongArray(N);
			for(int i = N - 1; i >= 0; i--) {
				res.addFirst(new Block(a[i], 1));
				collapse();
			}
			while(!res.isEmpty()) {
				Block cur = res.pollFirst();
				double val = (double) (cur.sum) / cur.vals;
				while(cur.vals --> 0) out.print(val + " ");
			}
			out.println();
		}
		
		class Block {
			long sum; int vals;
			Block(long ss, int vv) { sum = ss; vals = vv; }
			
			boolean shouldMerge(Block B) {
				// assume we are AHEAD of block B
				// current value is (sum / vals)
				// we can get (sum + B.sum) / (vals + B.vals)
				// let's merge if it's at least as good
				return (sum + B.sum) * vals <= sum * (vals + B.vals);
			}
			
			void merge(Block B) {
				sum += B.sum; vals += B.vals;
			}
			
			public String toString() {
				return String.format("(%d over %d)", sum, vals);
			}
		}

	}

	final static boolean cases = false;

	public static void main(String[] args) {

		FastScanner s = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		Solver solver = new Solver();
		for (int t = 1, T = cases ? s.nextInt() : 1; t <= T; t++)
			solver.solve(t, s, out);

		out.close();

	}

	static int min(int a, int b) {
		return a < b ? a : b;
	}

	static int max(int a, int b) {
		return a > b ? a : b;
	}

	static long min(long a, long b) {
		return a < b ? a : b;
	}

	static long max(long a, long b) {
		return a > b ? a : b;
	}

	static int swap(int a, int b) {
		return a;
	}

	static Object swap(Object a, Object b) {
		return a;
	}

	static String ts(Object... o) {
		return Arrays.deepToString(o);
	}

	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		public FastScanner(File f) throws FileNotFoundException {
			this(new FileInputStream(f));
		}

		public FastScanner(String s) {
			this.stream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
		}

		int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}

		// Jacob Garbage

		public int[] nextIntArray(int N) {
			int[] ret = new int[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextInt();
			return ret;
		}

		public int[][] next2DIntArray(int N, int M) {
			int[][] ret = new int[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextIntArray(M);
			return ret;
		}

		public long[] nextLongArray(int N) {
			long[] ret = new long[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextLong();
			return ret;
		}

		public long[][] next2DLongArray(int N, int M) {
			long[][] ret = new long[N][];
			for (int i = 0; i < N; i++)
				ret[i] = nextLongArray(M);
			return ret;
		}

		public double[] nextDoubleArray(int N) {
			double[] ret = new double[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDouble();
			return ret;
		}

		public double[][] next2DDoubleArray(int N, int M) {
			double[][] ret = new double[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDoubleArray(M);
			return ret;
		}

	}
}