Problem: Momiji has got a rooted tree, consisting of n nodes. The tree nodes are numbered by integers from 1 to n. The root has number 1. Momiji decided to play a game on this tree.

The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by v) and removes all the subtree nodes with the root in node v from the tree. Node v gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.

Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of nodes in the tree. The next n - 1 lines contain the tree edges. The i-th line contains integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the numbers of the nodes that are connected by the i-th edge.

It is guaranteed that the given graph is a tree.

Output

Print a single real number — the expectation of the number of steps in the described game.

The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Examples

Input

2
1 2


Output

1.50000000000000000000


Input

3
1 2
1 3


Output

2.00000000000000000000

Note

In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are: 

1 × (1 / 2) + 2 × (1 / 2) = 1.5

In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are: 

1 × (1 / 3) + (1 + 1.5) × (2 / 3) = (1 / 3) + (5 / 3) = 2

Solution (in JAVA):

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

public class TreeGame {

    List<List<Integer>> adjList;
    int [] d;
    double res;

    // Hah! I actually proved (after 1 day and asking a Ramsey-theory
    // professor for hints) that this works!
    void dfs(int curr, int par) {
        d[curr] = d[par]+1;
        res += 1.0/(d[curr] + 0.0);
        for (int i: adjList.get(curr))
            if (i != par)
                dfs(i, curr);
    }

    TreeGame() {
        Kattio io = new Kattio(System.in, System.out);
        int n = io.getInt();
        d = new int [n+1];
        adjList = new ArrayList<List<Integer>>(n+1);
        for (int i=0; i<= n; i++)
            adjList.add(new LinkedList<Integer>());
        for (int i=0; i<n-1; i++) {
            int a = io.getInt();
            int b = io.getInt();
            adjList.get(a).add(b);
            adjList.get(b).add(a);
        }
        dfs(1, 0);
        io.println(res);
        io.close();
    }


    public static void main(String [] argv) {
        new TreeGame();
    }

}


/** Simple yet moderately fast I/O routines.
 *
 * Example usage:
 *
 * Kattio io = new Kattio(System.in, System.out);
 *
 * while (io.hasMoreTokens()) {
 *    int n = io.getInt();
 *    double d = io.getDouble();
 *    double ans = d*n;
 *
 *    io.println("Answer: " + ans);
 * }
 *
 * io.close();
 *
 *
 * Some notes:
 *
 * - When done, you should always do io.close() or io.flush() on the
 *   Kattio-instance, otherwise, you may lose output.
 *
 * - 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();
    }



    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) { }
        return token;
    }

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