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.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class code6 {

	public static void main(String[] args) throws CloneNotSupportedException {
		// TODO Auto-generated method stub
		FastReader scn = new FastReader();
		int t = 1;
		//int p = t;
		while(t-->0) {
			int n = scn.nextInt();
			ArrayList<Integer>[] tree = new ArrayList[n];
			for(int i=0; i<n; i++)
				tree[i] = new ArrayList<Integer>();
			for(int i=0; i<n-1; i++) {
				int a = scn.nextInt()-1;
				int b = scn.nextInt()-1;
				tree[a].add(b);
				tree[b].add(a);
			}
			int a = 0;
			int u = -1;
			for(int i=0; i<n; i++) {
				int c = 0;
				for(int j=0; j<tree[i].size(); j++) {
					if(tree[tree[i].get(j)].size() == 1) {
						c++;
						if(u==-1)
							u=tree[i].get(j);
					}
				}
				if(c>0)
					c--;
				a += c;
			}
			
			int[] arr = new int[n];
			bfs(tree, arr, n, u);
			int flag = 0;
			for(int i=0; i<n; i++) {
				if(tree[i].size() == 1 && arr[i]%2 == 1) {
					flag = 1;
					break;
				}
			}
			
			if(flag == 1) 
				System.out.println("3 " + (n-1-a));
			else
				System.out.println("1 " + (n-1-a));
			
		}
			
		
	}
	
	public static void bfs (ArrayList<Integer>[] graph, int[] arr, int n, int i) {
		LinkedList<Integer> q = new LinkedList<>();
		//boolean[] vis = new boolean[n];
		for(int j=0; j<n; j++)
			arr[j] = Integer.MAX_VALUE;
		arr[i] = 0;
		q.addLast(i);
		while(!q.isEmpty()) {
			int u = q.removeFirst();
			//vis[u] = true;
			for(int j=0; j<graph[u].size(); j++) {
				if(arr[graph[u].get(j)] == Integer.MAX_VALUE) {
					arr[graph[u].get(j)] = arr[u]+1;
					q.addLast(graph[u].get(j));
				}
			}
		}
	}
	
	static int gcd(int a, int b) 
    { 
    if (a == 0) 
        return b;  
    return gcd(b % a, a);  
    } 
      
	//public static int 
    static int lcm(int a, int b) 
    { 
        return (a*b)/gcd(a, b); 
    } 
    
    static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}

	