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.*;
import java.text.*;

public class B1338 {

	static ArrayList<Integer>[] adjList;

	public static int solve(int u, int p) {
		if (adjList[u].size() == 1)
			return 2;
		int ans = 0;
		for (int x : adjList[u]) {
			if (x != p) {
				int s = solve(x, u);
				ans |= s == 3 ? 3 : s == 1 ? 2 : 1;
			}
		}
		return ans;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		adjList = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			adjList[i] = new ArrayList<Integer>();
		}
		for (int i = 0; i < n - 1; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			adjList[u].add(v);
			adjList[v].add(u);
		}
		int leaf = 0;
		int conn = 0;
		for (ArrayList<Integer> x : adjList) {
			if (x.size() == 1)
				leaf++;
			boolean h = false;
			for (int y : x) {
				if (adjList[y].size() == 1) {
					h = true;
				}
			}
			if (h)
				conn++;
		}
		int s = -1;
		for (int i = 0; i < n; i++) {
			if (adjList[i].size() != 1) {
				s = solve(i, -1);
				break;
			}
		}
		int max = (n - 1) - leaf + conn;
		int min = s == 2 || s == 1 ? 1 : 3;
		pw.println(min + " " + max);
		pw.close();

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < arr.length; i++)
				arr[i] = nextInt();
			return arr;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}
