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.util.*;
import java.math.*;
import java.io.*;
public class EdgeAssign {
	static int[] depth;
	static Map<Integer, ArrayList<Integer>> graph;
	static Set<Integer> leaves = new HashSet<Integer>();
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int n = Integer.parseInt(bf.readLine());
 		graph = new HashMap<Integer,ArrayList<Integer>>();
 		depth = new int[n+1];

 		for(int i = 1;i<=n;i++)
 			graph.put(i,  new ArrayList<Integer>());
 		for(int i = 0;i<n-1;i++){
 			StringTokenizer st = new StringTokenizer(bf.readLine());
 			int v1 = Integer.parseInt(st.nextToken());
 			int v2 = Integer.parseInt(st.nextToken());
 			graph.get(v1).add(v2);
 			graph.get(v2).add(v1);
 		}
 		int leaf = -1;
 		for(int i = 1;i<=n;i++)
 			if (graph.get(i).size() == 1)
 				leaf = i;
 		depth[leaf] = 0;
 		minDfs(leaf, -1);
 		boolean found = true;
 		for(int i : leaves){
 			if (depth[i]%2 == 1){
 				found = false;
 				break;
 			}
 		}
 		int min = 0;
 		if (found== true)
 			min = 1;
 		else
 			min = 3;
 		leaves.add(leaf);
 		int e = n-1;
 		int l = leaves.size();
 		
 		int m = 0;
 		for(int i =1;i<=n;i++){
 			for(int j : graph.get(i)){
 				if (leaves.contains(j)){
 					m++;
 					break;
 				}
 			}
 		}
 		out.println(min + " " + (e-l+m));
 		
 		
 		
 		out.close();
 		
 		
 		
 	}
	public static void minDfs(int v, int p){
		int count = 0;
		for(int child: graph.get(v)){
			if (child == p)
				continue;
			count++;
			depth[child] = depth[v]+1;
			minDfs(child, v);
		}
		if (count == 0)
			leaves.add(v);
	}
	
}
 	
 
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is

 		
 		
 		
 		
	


