You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:

  * For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0. 



Note that you can put very large positive integers (like 10^{(10^{10})}).

It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.

<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).

<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0. 

What are the minimum and the maximum possible values of f for the given tree? Find and print both.

Input

The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.

The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.

Output

Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.

Examples

Input


6
1 3
2 3
3 4
4 5
5 6


Output


1 4


Input


6
1 3
2 3
3 4
4 5
4 6


Output


3 3


Input


7
1 2
2 7
3 4
4 7
5 6
6 7


Output


1 6

Note

In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3. 

<image>

In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Andy Phan
 */
public class b {

    int MAX = 1_000_001;
    int MAX_LOG = 22;
    
    public static void main(String[] args) {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int n = in.nextInt();
        
        ArrayList<Integer>[] adj = new ArrayList[n];
        for(int i = 0; i < n; i++) adj[i] = new ArrayList<>();
        
        for(int i = 1; i < n; i++) {
            int u = in.nextInt()-1;
            int v = in.nextInt()-1;
            adj[u].add(v);
            adj[v].add(u);
        }
        
        int root = 0;
        for(int i = 0; i < n; i++) if(adj[i].size() == 1) root = i;
        
        ArrayDeque<Integer> q = new ArrayDeque<>();
        
        q.add(root);
        int[] depth = new int[n];
        
        depth[root] = 1;
        int num = n-1;
        for(int i = 0; i < n-1; i++) {
            int curr = q.poll();
            int tot = 0;
            for(int j : adj[curr]) {
                if(adj[j].size() == 1) tot++;
                if(depth[j] != 0) continue;
                depth[j] = depth[curr]+1;
                q.add(j);
            }
            if(tot > 0) num -= tot-1;
        }
        
        boolean has = false;
        for(int i = 0; i < n; i++) if(adj[i].size() == 1 && depth[i]%2 == 0) has = true;
        if(has) out.print(3 + " ");
        else out.print(1 + " ");
        out.println(num);
        
        out.close();
    }

    static class FS {

        BufferedReader in;
        StringTokenizer token;

        public FS(InputStream str) {
            in = new BufferedReader(new InputStreamReader(str));
        }

        public String nextLine() {
            token = new StringTokenizer("");
            try {
                return in.readLine();
            } catch (Exception e) {
                return null;
            }
        }

        public String next() {
            if (token == null || !token.hasMoreElements()) {
                try {
                    token = new StringTokenizer(in.readLine());
                } catch (IOException ex) {
                }
                return next();
            }
            return token.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}
