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.*;
import java.math.*;
import java.awt.Point;
 
public class Main {
	//static final long MOD = 998244353L;
	//static final long INF = 1000000000000000007L;
	static final long MOD = 1000000007L;
	static final int INF = 1000000007;
	
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		
		int N = sc.ni();
		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 n1 = sc.ni()-1;
			int n2 = sc.ni()-1;
			tree[n1].add(n2);
			tree[n2].add(n1);
		}
		
		int[] p = new int[N];
		int leaf = -1;
		for (int i = 0; i < N; i++) {
			if (tree[i].size() == 1) {
				p[tree[i].get(0)] += 1;
				leaf = i;
			}
		}
		
		int[] vis = new int[N]; //1 is odd, 2 is even
		vis[0] = 2;
		ArrayDeque<Integer> bfs = new ArrayDeque<Integer>();
		bfs.add(0);
		while (!bfs.isEmpty()) {
			int node = bfs.poll();
			for (int next: tree[node]) {
				if (vis[next]==0) {
					vis[next] = 3-vis[node];
					bfs.add(next);
				}
			}
		}
		
		int par = vis[leaf];
		for (int i = 0; i < N; i++) {
			if (tree[i].size() == 1) {
				//leaf
				if (vis[i] != par) {
					par = -1;
					break;
				}
			}
		}
		
		int min = 1;
		if (par == -1) {
			//there are two leaves in different components
			min = 3;
		}
		
		int max = (N-1);
		for (int i = 0; i < N; i++) {
			if (p[i] > 0) {
				//two edges are forced to be equal.
				max -= (p[i]-1);
			}
		}
		
		pw.println(min + " " + max);
		
		pw.close();
	}
 
 
	public static long dist(long[] p1, long[] p2) {
		return (Math.abs(p2[0]-p1[0])+Math.abs(p2[1]-p1[1]));
	}
	
	//Find the GCD of two numbers
	public static long gcd(long a, long b) {
		if (a < b) return gcd(b,a);
		if (b == 0)
			return a;
		else
			return gcd(b,a%b);
	}
	
	//Fast exponentiation (x^y mod m)
	public static long power(long x, long y, long m) { 
		if (y < 0) return 0L;
		long ans = 1;
		x %= m;
		while (y > 0) { 
			if(y % 2 == 1) 
				ans = (ans * x) % m; 
			y /= 2;  
			x = (x * x) % m;
		} 
		return ans; 
	}
	
	public static int[] shuffle(int[] array) {
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		return array;
	}
	
	public static long[] shuffle(long[] array) {
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    long temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		return array;
	}
	
	public static int[][] shuffle(int[][] array) {
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		return array;
	}
	
    public static int[][] sort(int[][] array) {
    	//Sort an array (immune to quicksort TLE)
 
		Arrays.sort(array, new Comparator<int[]>() {
			  @Override
        	  public int compare(int[] a, int[] b) {
				  return a[1]-b[1]; //ascending order
	          }
		});
		return array;
	}
    
    public static long[][] sort(long[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    long[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<long[]>() {
			  @Override
        	  public int compare(long[] a, long[] b) {
				  if (a[0] < b[0])
					  return -1;
				  else if (a[0] > b[0])
					  return 1;
				  else
					  return 0;
	          }
		});
		return array;
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            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 ni() { 
            return Integer.parseInt(next()); 
        }
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() {
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str;
        }
    }
}