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.*; 

public class Main {
	static int[] height;
	static int[] different;
	static boolean[] visited;
	static boolean one=false;
	static void dfs(ArrayList<ArrayList<Integer>> graph,int curr) {
//		System.out.println(curr+1+" yo");
		visited[curr]=true;
		boolean even=false;
		boolean odd=false;
		if(graph.get(curr).size()==1) {
			height[curr]=1;
		}
		else {
			int temp=0;
			for(int i=0;i<graph.get(curr).size();i++) {
				int child=graph.get(curr).get(i);
				if(!visited[child]) {
					dfs(graph,child);
					height[curr]=Math.max(height[curr], 1+height[child]);
					if(height[child]%2==0) {
						even=true;
					}
					else {
						odd=true;
					}
					if(height[child]==1) {
						temp=1;
					}
					else {
						different[curr]+=different[child]+1;
					}
				}
//				System.out.println(curr+" "+height[curr]+ " heih "+child);
				
			}
			if(even&& odd) {
				one=true;
			}
			different[curr]+=temp;
		}
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
//		t=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		while(t>0) {
			t--;
			int n=f.nextInt();
			ArrayList<ArrayList<Integer>> graph=new ArrayList<>();
			for(int i=0;i<n;i++) {
				graph.add(new ArrayList<>());
			}
			for(int i=0;i<n-1;i++) {
				int u=f.nextInt()-1;
				int v=f.nextInt()-1;
				graph.get(u).add(v);
				graph.get(v).add(u);
//				System.out.println(graph);
			}
//			System.out.println(graph);
			int u=-1;
			for(int i=0;i<graph.size();i++) {
				if(graph.get(i).size()>1) {
					u=i;
					break;
				}
			}
			height=new int[n];
			different=new int[n];
			visited=new boolean[n];
//			System.out.println(u);
			dfs(graph,u);
			int ans=different[u];
			if(one) {
				System.out.println(3+" "+ans);
			}
			else {
				System.out.println(1+" "+ans);
			}
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	