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.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=998244353L;
	long inf=Long.MAX_VALUE;
	long gcd(long a,long b) {
		return a==0?b:gcd(b%a,a);
	}
	ArrayList<Integer>[] graph;
	int[] degree;
	int max,min=1;
	void work() {
		int n=ni();
		max=n-1;
		degree=new int[n];
		graph=ng(n,n-1);
		int start=0;
		for(int i=0;i<n;i++) {
			if(degree[i]==1) {
				start=i;
				break;
			}
		}
		int[] A=dfs(start,-1);
		if(A[1]>0) {
			min=3;
		}
		out.println(min+" "+max);
	}
	
	
	private int[] dfs(int node, int pre) {
		int[] ret=new int[2];
		int c=0;
		for(int nn:graph[node]) {
			if(degree[nn]==1) {
				c++;
			}
			if(nn!=pre) {
				int[] r=dfs(nn,node);
				ret[0]+=r[1];
				ret[1]+=r[0];
			}
		}
		if(degree[node]==1) {
			ret[0]=1;
		}
		max-=Math.max(0, c-1);
		return ret;
	}


	//input
	@SuppressWarnings("unused")
	private ArrayList<Integer>[] ng(int n, int m) {
		ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			int s=in.nextInt()-1,e=in.nextInt()-1;
			graph[s].add(e);
			graph[e].add(s);
			degree[s]++;
			degree[e]++;
		}
		return graph;
	}

	private ArrayList<long[]>[] ngw(int n, int m) {
		ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
			graph[(int)s].add(new long[] {e,w,i});
			graph[(int)e].add(new long[] {s,w});
		}
		return graph;
	}

	private int ni() {
		return in.nextInt();
	}

	private long nl() {
		return in.nextLong();
	}
	
	private String ns() {
		return in.next();
	}

	private long[] na(int n) {
		long[] A=new long[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextLong();
		}
		return A;
	}
	private int[] nia(int n) {
		int[] A=new int[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextInt();
		}
		return A;
	}
}	

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

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


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

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

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