Problem: Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. Cities and roads of USC form a rooted tree (Barney's not sure why it is rooted). Root of the tree is the city number 1. Thus if one will start his journey from city 1, he can visit any city he wants by following roads.

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Some girl has stolen Barney's heart, and Barney wants to find her. He starts looking for in the root of the tree and (since he is Barney Stinson not a random guy), he uses a random DFS to search in the cities. A pseudo code of this algorithm is as follows:
    
    
      
    let starting_time be an array of length n  
    current_time = 0  
    dfs(v):  
    	current_time = current_time + 1  
    	starting_time[v] = current_time  
    	shuffle children[v] randomly (each permutation with equal possibility)  
    	// children[v] is vector of children cities of city v  
    	for u in children[v]:  
    		dfs(u)  
    

As told before, Barney will start his journey in the root of the tree (equivalent to call dfs(1)).

Now Barney needs to pack a backpack and so he wants to know more about his upcoming journey: for every city i, Barney wants to know the expected value of starting_time[i]. He's a friend of Jon Snow and knows nothing, that's why he asked for your help.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 105) — the number of cities in USC.

The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi < i), where pi is the number of the parent city of city number i in the tree, meaning there is a road between cities numbered pi and i in USC.

Output

In the first and only line of output print n numbers, where i-th number is the expected value of starting_time[i].

Your answer for each city will be considered correct if its absolute or relative error does not exceed 10 - 6.

Examples

Input

7
1 2 1 1 4 4


Output

1.0 4.0 5.0 3.5 4.5 5.0 5.0 


Input

12
1 1 2 2 4 4 3 3 1 10 8


Output

1.0 5.0 5.5 6.5 7.5 8.0 8.0 7.0 7.5 6.5 7.5 8.0 

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.List;
import java.util.NoSuchElementException;

public class Main {

	private static Graph g;
	private static int[] num;
	private static double[] ans;

	private static int dfs(int now) {
		int ret = 1;
		for (Graph.Path p : g.get(now)) {
			ret += dfs(p.to);
		}
		num[now] = ret;
		return ret;
	}

	private static void dfs2(int now) {

		for (Graph.Path p : g.get(now)) {
			ans[p.to] = ans[now] + (double) (1.0 + (num[now] - num[p.to] - 1) / 2.0);
			dfs2(p.to);
		}
	}

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		num = new int[n + 1];
		ans = new double[n + 1];
		g = new Graph(n + 1);
		for (int i = 2; i <= n; i++) {
			g.add(sc.nextInt(), i);
		}

		dfs(1);
		ans[1] = 1;
		dfs2(1);

		StringBuilder sb = new StringBuilder();
		for (int i = 1; i <= n; i++) {
			sb.append(ans[i]);
			sb.append(" ");
		}
		System.out.println(sb.substring(0, sb.length() - 1));
	}

}

class Graph {
	public static class Path implements Comparable<Path> {
		public final int from;
		public final int to;
		public final long cost;

		public Path(int from, int to, long cost) {
			this.from = from;
			this.to = to;
			this.cost = cost;
		}

		@Override
		public int compareTo(Path o) {
			return Long.compare(this.cost, o.cost);
		}
	}

	private final List<List<Path>> edgeList;
	private final int size;

	public Graph(int n) {
		this.size = n;
		this.edgeList = new ArrayList<List<Path>>();
		for (int i = 0; i < n; i++) {
			this.edgeList.add(new ArrayList<Path>());
		}
	}

	public void add(int from, int to, long cost) {
		this.edgeList.get(from).add(new Path(from, to, cost));
	}

	public void addBoth(int v1, int v2, long cost) {
		this.add(v1, v2, cost);
		this.add(v2, v1, cost);
	}

	public void add(int from, int to) {
		this.edgeList.get(from).add(new Path(from, to, 1));
	}

	public void addBoth(int v1, int v2) {
		this.add(v1, v2, 1);
		this.add(v2, v1, 1);
	}

	public List<Path> get(int from) {
		return this.edgeList.get(from);
	}

	public int size() {
		return this.size;
	}
}

class FastScanner {
	public static String debug = null;

	private final InputStream in = System.in;
	private int ptr = 0;
	private int buflen = 0;
	private byte[] buffer = new byte[1024];
	private boolean eos = false;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				if (debug != null) {
					buflen = debug.length();
					buffer = debug.getBytes();
					debug = "";
					eos = true;
				} else {
					buflen = in.read(buffer);
				}
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen < 0) {
				eos = true;
				return false;
			} else if (buflen == 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean isEOS() {
		return this.eos;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		return (int) nextLong();
	}

	public long[] nextLongList(int n) {
		return nextLongTable(1, n)[0];
	}

	public int[] nextIntList(int n) {
		return nextIntTable(1, n)[0];
	}

	public long[][] nextLongTable(int n, int m) {
		long[][] ret = new long[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				ret[i][j] = nextLong();
			}
		}
		return ret;
	}

	public int[][] nextIntTable(int n, int m) {
		int[][] ret = new int[n][m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				ret[i][j] = nextInt();
			}
		}
		return ret;
	}
}