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 ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — 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.
题目分析：
看到城市之间走路就觉得这题可能用深搜，既然思路出来了，就先存储地图呗，但当我敲完所有代码的时候，一构建，发现数组过大，这我也很懵逼。。从来没遇到过，一查原来一维数组只能开1e8，二维是1e4，但题目范围刚好是用到二维1e5，再一查，原来还可以用动态数组，定义一个二级指针，再new一个，后来发现还可以用vector。这也算是知识盲区了，以前完全不知道。
代码：
#include<iostream>
#include<iomanip>
#include<vector>
using namespace std;
int v[100050];//存储有没有到达过
vector<int>a[100050];//动态数组
double dfs(int u)//表示从u城市开始所得到的期望
{
v[u]=1;
double ans=0,k=0;
int len=a[u].size();
for(int i=0;i<len;i++){
if(v[a[u][i]]==0){//没到过这个城市
v[a[u][i]]=1;
ans+=dfs(a[u][i]);
k++;//每返回一个值就加一，代表终点城市增加一个
}
}
if(k!=0) ans/=k;//当经过的城市大于等于1个时求出期望
if(u!=1) ans++;
return ans;
}
int main()
{
int n,c,b;
cin>>n;
for(int i=0;i<n-1;i++){
cin>>b>>c;
a[b].push_back(c);
a[c].push_back(b);
}
cout<<fixed<<setprecision(15)<<dfs(1);
return 0;
}
