D. Puzzles(Codeforces Round #362 (Div. 2))

D. Puzzles

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, ..., p_n (1 ≤ p_i < i), where p_i is the number of the parent city of city number i in the tree, meaning there is a road between cities numbered p_i 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 

题意：给你一棵树，然后用dfs随即给每个点标号，求每个点标号的期望；
思路：dfs+概率dp；
我们可以知道根节点的期望为1；
然后，他的子节点都是等概率的。
dp数组是各个节点的期望。
先给样例一的图:

我们写第二层的排列
1，1 2 4 5
2，1 2 5 4
3，1 4 2 5
4，1 4 5 2
5，1 5 2 4
6，1 5 4 2
所以节点2的期望为dp[1]+（(1+size(4)+size(5)）*2+size（4）+1+size(5)+1+1+1）/6；
根据这个我们先试着猜想dp[v]=dp[u]+1+(size(u)-size(v)-1)/2;
这就是状态转移方程；
我们先把上面到下面一层所必须加上一步先加上，也就是dp[u]+1;
然后我们可以知道下面所有的排列中，此节点的兄弟节点，要么排在这个节点之前要么之后，所一其他节点对于该节点的贡献为size(）/2；
也就是排在前面和后面是等概率的；
复杂度O（n）

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<stdlib.h>
#include<vector>
using namespace std;
vector<int>vec[100006];
long long cnt[100006];
long long dfs1(int n);
double  dp[100006];
void dfs(int n);
int main(void)
{
        int i,j,k;
        while(scanf("%d",&k)!=EOF)
        {
                int n;memset(dp,0,sizeof(dp));
                for(i=0; i<100006; i++)
                {
                        cnt[i]=0;
                        vec[i].clear();
                }
                for(i=2; i<=k; i++)
                {
                        scanf("%d",&n);
                        vec[n].push_back(i);
                }
                long long t=dfs1(1);
                dp[1]=1.0;
                dfs(1);printf("%.1f",dp[1]);
                for(i=2; i<=k; i++)
                {
                        printf(" %.1f",dp[i]);
                }
                printf("
");
        }
        return 0;
}
long long dfs1(int n)
{
        long long sum;
        int i,j,k;
        for(i=0; i<vec[n].size(); i++)
        {
                cnt[n]+=dfs1(vec[n][i]);
        }
        cnt[n]+=1;
        return cnt[n];
}
void dfs(int n)
{
        int i,j;
        for(i=0; i<vec[n].size(); i++)
        {
                int x;
                x=vec[n][i];
                dp[x]=dp[n]+1.0;
                dp[x]+=1.0*(cnt[n]-cnt[x]-1)/2.0;
                dfs(x);
        }
}