C. Journey

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 ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by thei-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 .

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 ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by thei-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.

题意：给你一棵树，每一条边的权值为1，让你求从1到叶子节点边权和的期望，到达每一个相邻节点的概率是一样的，不能走已走过的边。

题解：

dfs做一遍树形dp。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
double ans;
struct node
{
    int next,to;
}edge[200005];
int head[100001],size=1;
void putin(int from,int to)
{
    size++;
    edge[size].to=to;
    edge[size].next=head[from];
    head[from]=size;
}
void dfs(int u,int fa,double s,int depth)
{
    int i,cnt=0;
    for(i=head[u];i!=-1;i=edge[i].next)if(edge[i].to!=fa)cnt++;
    if(cnt==0){ans+=s*(double)depth;return;}
    for(i=head[u];i!=-1;i=edge[i].next)
        if(edge[i].to!=fa)
            dfs(edge[i].to,u,s*(double)(1/(double)cnt),depth+1);
}
int main()
{
    int i,j;
    memset(head,-1,sizeof(head));
    scanf("%d",&n);
    for(i=1;i<n;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        putin(a,b);
        putin(b,a);
    }
    dfs(1,0,1,0);
    printf("%.15lf",ans);
    return 0;
}