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 Main{
	static LinkedList<Integer>[]adj;
	static int max;
	static boolean odd=false;
	static void dfs(int i,int par,int depth) {
		if(adj[i].size()==1 && depth%2==1) {
			odd=true;
		}
		for(int j:adj[i]) {
			if(j==par)continue;
			dfs(j, i, depth+1);
		}
	}
	
	static boolean[]leaf;
	static void dfs2(int i,int par) {
		int cnt=0;
		for(int j:adj[i]) {
			if(j==par)continue;
			dfs2(j, i);
			if(leaf[j])cnt++;
		}
		if(cnt>1) {
			max-=cnt;
			max++;
		}
	}
	public static void main(String[] args) throws Exception{
		PrintWriter pw=new PrintWriter(System.out);
		MScanner sc = new MScanner(System.in);
		int n=sc.nextInt();
		adj=new LinkedList[n];
		for(int i=0;i<n;i++)adj[i]=new LinkedList<>();
		for(int i=0;i<n-1;i++) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1;
			adj[x].add(y);
			adj[y].add(x);
			
		}
		for(int i=0;i<n;i++) {
			if(adj[i].size()==1) {
				dfs(i, -1, 0);
				break;
			}
		}
		leaf=new boolean[n];
		max=n-1;
		for(int i=0;i<n;i++) {
			if(adj[i].size()==1) {
				leaf[i]=true;
			}
		}
		for(int i=0;i<n;i++) {
			if(!leaf[i]) {
				dfs2(i, -1);
				break;
			}
		}
//		pw.println(max);
		if(!odd) {
			pw.println(1+" "+max);
			
			
		}
		else
			pw.print(3+" "+max);
		
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		static void shuffle(int[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				int tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		static void shuffle(long[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				long tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void addX(int[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
	static void addX(long[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
}