Codeforces Round #362 (Div. 1) B. Puzzles(递推)

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本文介绍了一道关于在有根树中使用随机深度优先搜索计算各节点时间戳期望值的问题，通过递归计算子节点数量并利用子节点与兄弟节点间的关系确定每个节点的时间戳期望。

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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 number1. Thus if one will start his journey from city 1, he can visit any city he wants by following roads.

$\text{[math]}$

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 arandom 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 calldfs(1)).

Now Barney needs to pack a backpack and so he wants to know more about his upcoming journey: for every cityi, Barney wants to know the expected value ofstarting_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 integersp₂, p₃, ..., p_n (1 ≤ p_i < i), where p_i is the number of the parent city of city numberi in the tree, meaning there is a road between cities numberedp_i andi in USC.

Output

In the first and only line of output print n numbers, wherei-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 exceed10^- 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

题意：给定一棵有根树，问每个节点的时间戳的期望值。

分析：很容易想到子节点的期望可以根据父节点得出，然后考虑兄弟对子节点的影响，每个兄弟在其前后的概率相同都是0.5,所以节点v的期望为(Size[u]-Size[v]-1)/2+1+f[u].

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<queue>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MOD 1000000007
#define MAXN 200005
using namespace std;
int n,Size[MAXN];
vector<int> G[MAXN];
double ans[MAXN];
int dfs(int u)
{
    Size[u] = 1;
    for(int v : G[u]) Size[u] += dfs(v);
    return Size[u];
}
void dfs2(int u,double val)
{
    ans[u] = val;
    for(int v : G[u])
     dfs2(v,val + (Size[u]-Size[v]-1)*0.5 + 1);
}
int main()
{
    scanf("%d",&n);
    for(int i = 2;i <= n;i++)
    {
        int u;
        scanf("%d",&u);
        G[u].push_back(i);
    }
    dfs(1);
    dfs2(1,1);
    for(int i = 1;i <= n;i++) printf("%.7f ",ans[i]);
}