Problem: 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 <image>.

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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;
import java.util.concurrent.*;

public final class round_428_c
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static ArrayList<Integer>[] al;
	static boolean[] leaf;
	static double[] level,prob;
	
	static void dfs(int u,int p,double curr,double xx)
	{
		level[u]=curr;double ctr=0;
		
		prob[u]=xx;
		
		for(int x:al[u])
		{
			if(x!=p)
			{
				ctr++;
			}
		}
		
		double mul=1.0/ctr;
		
		for(int x:al[u])
		{
			if(x!=p)
			{
				dfs(x,u,curr+1,xx*mul);
			}
		}
		
		leaf[u]=(ctr==0);
	}
	
	@SuppressWarnings("unchecked")
    public static void main(String args[]) throws Exception
    {
		int n=sc.nextInt();al=new ArrayList[n];
		
		for(int i=0;i<n;i++)
		{
			al[i]=new ArrayList<Integer>();
		}
		
		for(int i=1;i<n;i++)
		{
			int u=sc.nextInt()-1,v=sc.nextInt()-1;
			
			al[u].add(v);al[v].add(u);
		}
		
		level=new double[n];leaf=new boolean[n];prob=new double[n];dfs(0,-1,0,1.0);
		
		double res=0;
		
		for(int i=0;i<n;i++)
		{
			if(leaf[i])
			{	
				double curr=(prob[i]*level[i]);
				
				res+=curr;
			}
		}
		
		out.printf("%.20f\n",res);out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}