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.

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 

題意：到達某一個節點，下一個訪問的子代概率相同，按照dfs的順序，求每個節點被訪問到的時間

題解：對於對某個節點，設其父節點有另一分支a ，則該節點要麼在a之前被訪問，要麼在a被全部訪問完後，才被訪問。同樣我們把其父節點其他所有分支看成一個整體，該節點要麼在這個整體之前要麼在他之後，所以可得

該節點的期望就為： 其他分支的節點和 / 2  +  1 + 父節點的期望

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
const int N=1e5+10;
vector<int> v[N];
int n,son[N];
double dp[N];
void dfs1(int u)
{
	son[u]=1;
	for(int i=0;i<v[u].size();i++)
	{
		int to=v[u][i];
		dfs1(to);
		son[u]+=son[to];
	}
}
void dfs2(int u)
{
	for(int i=0;i<v[u].size();i++)
	{
		int to=v[u][i];
		dp[to]=dp[u]+(double)(son[u]-son[to]-1)/2+1;
		dfs2(to);
	}
}
int main()
{
	int fa;
	scanf("%d",&n);
	for(int i=2;i<=n;i++)
	{
		scanf("%d",&fa);
		v[fa].pb(i);
	}
	dfs1(1);
	dp[1]=1;
	dfs2(1);
	for(int i=1;i<=n;i++)printf("%.10f%c",dp[i]," \n"[i==n]);
	return 0;
}