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 static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.Math.PI;
import static java.lang.Math.min;
import static java.lang.System.arraycopy;
import static java.lang.System.exit;
import static java.util.Arrays.copyOf;

import java.util.LinkedList;
import java.util.Iterator;
import java.io.FileReader;
import java.io.FileWriter;
import java.math.BigInteger; 
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Comparator;
import java.lang.StringBuilder;
import java.util.Collections;
public class Solution {

	static int scanInt() throws IOException {
		return parseInt(scanString());
	}

	static long scanLong() throws IOException {
		return parseLong(scanString());
	}
	static double scanDouble() throws IOException {
		return parseDouble(scanString());
	}

	static String scanString() throws IOException {
		if (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}
	static String scanLine() throws IOException {
		return in.readLine();
	}


	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;
	static LinkedList<Integer> edges[]; static int depth[], Max_N, wasted, least;
	
	public static void main(String[] args) {
		try {
			long startTime = System.currentTimeMillis();

			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			//in = new BufferedReader(new FileReader("input.txt"));
			//out = new PrintWriter(new FileWriter("output.txt"));
			Max_N=(int)1e5+1;
			edges= new LinkedList[Max_N];
			depth= new int[Max_N];
			least=1;
			solve();
			
			long endTime   = System.currentTimeMillis();
			long totalTime = endTime - startTime;
			//System.out.println(totalTime+"  "+System.currentTimeMillis() );
			in.close();
			out.close();
			
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
	private static void solve() throws IOException{
		int n=scanInt();
		for(int i=0; i<=n; ++i){
			edges[i]=new LinkedList<>();
		}
		for(int i=0; i<n-1; ++i){
			int a=scanInt(), b=scanInt();
			edges[a].add(b);
			edges[b].add(a);
		}
		for(int i=1; i<=n; ++i){
			if(isLeaf(i)){
				dfs(i, -1);
				break;
			}
			
		}
		int atmost= n-wasted-1;
		out.println(least+" "+atmost);
		
	}
	private static void dfs(int node, int parent){
		if(isLeaf(node) && depth[node]%2==1){
			least=3;
		}
		int leaves=0;
		for(int x : edges[node]){
			if(isLeaf(x))
				++leaves;
			if(x==parent)
				continue;
			depth[x]=depth[node]+1;
			dfs(x, node);
		}
		if(leaves>=2){
			wasted+=leaves-1;
		}
	}
	private static boolean isLeaf(int node){
		return edges[node].size()==1;
	}

	
	
	
	
	
	

	
	
	
}
 

