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.*;





public class Solution {
    
    private class Node{
    int id;
    double pathlen = 0.00;
    double probability;
    boolean visited;
    boolean isRoot;
    ArrayList<Node> connectedTo = new ArrayList<>();

        Node(int id) {
            this.id = id;
        }
    }

    Kattio io = new Kattio(System.in, System.out);
    int n;
    Node[] nodes;


    private void dfs(Node parent, Node curr){
        curr.visited = true;
        double childCount = getChildCount(parent, curr);
        if(childCount == 0){
            curr.isRoot = true;
            return;
        }
        //if current node is root
        //traverse all children of current node and set their length and probability
        for(Node towards : curr.connectedTo){
            if(towards != parent){
                towards.pathlen = curr.pathlen+1.00;
                towards.probability = curr.probability * (1.00/childCount);
                dfs(curr, towards);
            }
        }
    }

    private double getChildCount(Node parent, Node curr) {
        if(parent == null){
            return (double) curr.connectedTo.size();
        }
        return (double) curr.connectedTo.size()-1;
    }


    Solution() {
        n = io.getInt();
        nodes = new Node[n+1];
        for(int i=1; i<=n; i++){
            nodes[i] = new Node(i);
        }
        for(int i=0; i<n-1; i++){
            int from = io.getInt();
            int to = io.getInt();
            nodes[to].connectedTo.add(nodes[from]);
            nodes[from].connectedTo.add(nodes[to]);
        }
        nodes[1].probability = 1.00;
        dfs(null, nodes[1]);
        double res = 0.00;
        for(int i=1; i<=n; i++){
            if(nodes[i].isRoot){
                res += (nodes[i].pathlen * nodes[i].probability);
            }
        }
        io.println(res);
        io.close();
    }

    public static void main(String[] args) {
        new Solution();
    }
}



/**
 * Simple yet moderately fast I/O routines.
 * <p>
 * Example usage:
 * <p>
 * Kattio io = new Kattio(System.in, System.out);
 * <p>
 * while (io.hasMoreTokens()) {
 * int n = io.getInt();
 * double d = io.getDouble();
 * double ans = d*n;
 * <p>
 * io.println("Answer: " + ans);
 * }
 * <p>
 * io.close();
 * <p>
 * <p>
 * Some notes:
 * <p>
 * - When done, you should always do io.close() or io.flush() on the
 * Kattio-instance, otherwise, you may lose output.
 * <p>
 * - The getInt(), getDouble(), and getLong() methods will throw an
 * exception if there is no more data in the input, so it is generally
 * a good idea to use hasMoreTokens() to check for end-of-file.
 *
 * @author: Kattis
 */

class Kattio extends PrintWriter {
    public Kattio(InputStream i) {
        super(new BufferedOutputStream(System.out));
        r = new BufferedReader(new InputStreamReader(i));
    }

    public Kattio(InputStream i, OutputStream o) {
        super(new BufferedOutputStream(o));
        r = new BufferedReader(new InputStreamReader(i));
    }

    public boolean hasMoreTokens() {
        return peekToken() != null;
    }

    public int getInt() {
        return Integer.parseInt(nextToken());
    }

    public double getDouble() {
        return Double.parseDouble(nextToken());
    }

    public long getLong() {
        return Long.parseLong(nextToken());
    }

    public String getWord() {
        return nextToken();
    }

    public String getLine() {
        try {
            return r.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }


    private BufferedReader r;
    private String line;
    private StringTokenizer st;
    private String token;

    private String peekToken() {
        if (token == null)
            try {
                while (st == null || !st.hasMoreTokens()) {
                    line = r.readLine();
                    if (line == null) return null;
                    st = new StringTokenizer(line);
                }
                token = st.nextToken();
            } catch (IOException e) {
                e.printStackTrace();
            }
        return token;
    }

    private String nextToken() {
        String ans = peekToken();
        token = null;
        return ans;
    }
}