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
41 21 32 4
Output
1.500000000000000
Input
51 21 33 42 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.
求数学期望，dfs做，因为n个点只有n-1个边，所以就没有连成圈，len的长度其实就时与1相连的个数，然后用ans来保持最底层第的深度和 答案就时ans/len了。
比如第二组数据，与1相连的点有两个，所以len是2，最底层第分别时4和5，深度分别2,2，所以答案就时(2+2)/len。
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int N = 100010;
4 vector<int> G[N];
5
6 double dfs(int v, int p) {
7 double ans = 0;
8 int len = 0;
9 for(int i = 0; i < G[v].size(); i ++) {
10 int u = G[v][i];
11 if(u == p) continue;
12 ans += (1 + dfs(u, v));
13 len ++;
14 }
15 return len == 0 ? 0.0 : ans / len;
16 }
17 int main() {
18 int n;
19 scanf("%d", &n);
20 for(int i = 1; i < n; i ++) {
21 int u, v;
22 scanf("%d %d", &u, &v);
23 G[u].push_back(v);
24 G[v].push_back(u);
25 }
26 printf("%.15f\n",dfs(1,0));
27 return 0;
28 }