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 sun.awt.image.ImageWatched;

import java.util.*;
import java.io.*;

public class B{
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(f.readLine());
        LinkedList<Integer>[] edges = new LinkedList[n+1];
        for(int i = 0; i <= n; i++){
            edges[i] = new LinkedList<Integer>();
        }
        for(int i = 0; i < n-1; i++){
            StringTokenizer st = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            edges[a].add(b);
            edges[b].add(a);
        }
        int root = -1;
        for(int i = 1; i <= n; i++){
            if(edges[i].size() > 1){
                root = i;
                break;
            }
        }
        HashSet<Integer> leaves = new HashSet<Integer>();
        for(int i = 1; i <= n; i++){
            if(edges[i].size() == 1) leaves.add(i);
        }
        ArrayList<Integer> ordering = new ArrayList<Integer>();
        HashSet<Integer> seen = new HashSet<Integer>();
        HashSet<Integer> actuallyseen = new HashSet<Integer>();
        Stack<Integer> stk = new Stack<Integer>();
        stk.add(root);
        while(!stk.isEmpty()){
            int temp = stk.pop();
            if(actuallyseen.contains(temp)) continue;
            if(seen.contains(temp)){
                ordering.add(temp);
                actuallyseen.add(temp);
                continue;
            }
            seen.add(temp);
            stk.add(temp);
            for(int e: edges[temp]){
                if(!seen.contains(e) && !actuallyseen.contains(e)){
                    stk.add(e);
                }
            }
        }
        int ans1 = 1;
        boolean[] evenpaths = new boolean[n+1];
        boolean[] oddpaths = new boolean[n+1];
        HashSet<Integer> seen2 = new HashSet<Integer>();
        for(int i: ordering){
            if(edges[i].size() == 1){
                evenpaths[i] = true;
                seen2.add(i);
            }else{
                boolean odd = false;
                boolean even = false;
                boolean done = false;
                for(int e: edges[i]){
                    if(seen2.contains(e)){
                        if((oddpaths[e] && even) || evenpaths[e] && odd){
                            ans1 = 3;
                            done = true;
                            break;
                        }else{
                            if(oddpaths[e]) odd = true;
                            if(evenpaths[e]) even = true;
                        }
                    }
                    if(oddpaths[e]) evenpaths[i] = true;
                    if(evenpaths[e]) oddpaths[i] = true;
                }
                seen2.add(i);
                if(done) break;
            }
        }
        int ans2 = n-1;
        for(int i = 1; i <= n; i++){
            int count = 0;
            for(int e: edges[i]){
                if(leaves.contains(e)) count++;
            }
            ans2-=(Math.max(0,count-1));
        }
        out.println(ans1 + " " + ans2);

        out.close();
    }
}
