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.*;
import java.util.*;
public class cp1 {
	static boolean visited[];
	static boolean chk=true;
	static int dfs(ArrayList<Integer> arr[], int i){
		int h=0,h1=-1;
		for(int j: arr[i]){
			if(!visited[j]){
				visited[j]=true;
				h=dfs(arr, j);
				if(h1!=-1 && (h+h1)%2!=0)
					chk=false;
				h1=h;
			}
		}
		return h+1;
	}
	static int calcMax(ArrayList<Integer> arr[], int n){
		int l=0, m=0;
		for(int i=1;i<=n;i++){
			if(arr[i].size()==1)
				l++;
			for(int j: arr[i]){
				if(arr[j].size()==1){
					m++;
					break;
				}
			}
		}
		// System.out.println((n-1)+" "+l+" "+m);
		return n-1-l+m;
	}
	static StringBuilder run(){
		StringBuilder sb=new StringBuilder();
		// code
		int n=in.nextInt();
		visited=new boolean[n+1];
		ArrayList<Integer> arr[]=new ArrayList[n+1];
		for(int i=1;i<=n;i++)
			arr[i]=new ArrayList<>();
		for(int i=0;i<n-1;i++){
			int x=in.nextInt();
			int y=in.nextInt();
			arr[x].add(y);
			arr[y].add(x);
		}
		dfs(arr, 1);
		int max=calcMax(arr, n);
		if(chk)
			sb.append(1+" ");
		else
			sb.append(3+" ");
		sb.append(max);
		return sb;
	}

	static StringBuilder solve() {
		StringBuilder result=new StringBuilder();
		int t=in.nextInt();
		while(--t>=0)
			result.append(run()+"\n");
		return result;
	}

	// Sankcan
	public static void main(String args[]) {System.out.print(run());}
	static scan in=new scan(System.in);
}
class scan
{
    public static BufferedReader reader;
    public static StringTokenizer token;

    scan(InputStream str)
    {
        reader=new BufferedReader(new InputStreamReader(str));
        token=null;
    }

    static int nextInt()
    {
        while(token==null||!token.hasMoreTokens())
        {
            try { token=new StringTokenizer(reader.readLine());   }
            catch(IOException e){  throw new RuntimeException(e); }
        }
        return Integer.parseInt(token.nextToken());
    }

    static long nextLong()
    {
        while(token==null||!token.hasMoreTokens())
        {
            try { token=new StringTokenizer(reader.readLine());   }
            catch(IOException e){  throw new RuntimeException(e); }
        }
        return Long.parseLong(token.nextToken());
    }

    static String next()
    {
        while(token==null||!token.hasMoreTokens())
        {
            try { token=new StringTokenizer(reader.readLine());   }
            catch(IOException e){  throw new RuntimeException(e); }
        }
        return token.nextToken();
    }

}
