Codeforces Round #428

C. Journey

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.

Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.

Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link https://en.wikipedia.org/wiki/Expected_value.

Input

The first line contains a single integer n (1 ≤ n ≤ 100000) — number of cities.

Then n - 1 lines follow. The i-th line of these lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — the cities connected by the i-th road.

It is guaranteed that one can reach any city from any other by the roads.

Output

Print a number — the expected length of their journey. The journey starts in the city 1.

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

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Examples

Input

4
1 2
1 3
2 4

Output

1.500000000000000

Input

5
1 2
1 3
3 4
2 5

Output

2.000000000000000

Note

In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.

In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.

题目大意：给定一个 n 个结点 n - 1 条边的连通图（其实就是一棵树）和一个出发点。每次只去没走过的结点，没有符合上述条件的点时就停下来。求这个人所有可能的路径长度的期望。

别被那个期望给唬住了，其实很简单，把图DFS一遍就行了，DSF的时候记下当前深度（即路径长度）和概率。一开始在1的概率为零，每次往下走的时候概率除以该点的出度。到头了往 ans 上加一下就行了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 1e5 + 5;
vector<int> v[maxn];
double ans = 0;
bool vis[maxn];
void DFS(int x,int d,double p)
{
    //printf("%d %f\n",x,p);
    int len = v[x].size();
    int tt = x == 1 ? len : len - 1;
    if(len == 1 && x != 1)
    {
        //printf("%d %d %f\n",x,d,p);
        ans += d * p;
        return;
    }
    for(int i = 0;i < len; ++i)
    {
        int u = v[x][i];
        if(!vis[u])
        {
            vis[u] = true;
            DFS(u,d + 1,p / tt);
        }
    }
}
int main()
{
    int n,x,y;
    scanf("%d",&n);
    for(int i = 1;i < n; ++i)
    {
        scanf("%d %d",&x,&y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    memset(vis,false,sizeof(vis));
    vis[1] = true;
    DFS(1,0,1);
    printf("%.15f\n",ans);
    return 0;
}