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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;


public class _Edge_Weight_Assignment {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			graph.add(new ArrayList<Integer>());
		}
		int[] deg = new int[n];
		for (int i = 0; i < n - 1; i++) {
			String[] sl = br.readLine().split(" ");
			int v = Integer.parseInt(sl[0]) - 1;
			int v2 = Integer.parseInt(sl[1]) - 1;
			deg[v]++;
			deg[v2]++;
			graph.get(v).add(v2);
			graph.get(v2).add(v);
		}
		int root = -1;
		for (int i = 0; i < n; i++) {
			if (deg[i] == 1) {
				root = i;
				break;
			}
		}

		
		
		int max = n - 1;
		int min=1;
		boolean vs=dfs(root, -1, graph, 0);
		if(vs){
			min=3;
		}
		for (int i = 0; i < n; i++) {
			int c = 0;
			for (int j : graph.get(i)) {
				if (graph.get(j).size() == 1) {
					c++;
				}
			}
			if(c>=2){
			max -= c - 1;}
			
		}
		System.out.println(min + " " + max);
	}


	public static boolean dfs(int v, int p, ArrayList<ArrayList<Integer>> graph,int ht) {
	boolean haschild=false;
	boolean vs=false;
		for (int child : graph.get(v)) {
			if (child == p) {
				continue;
			}
          haschild=true;
		vs=vs|| dfs(child, v, graph,ht+1);
		}
		if (ht%2==1&&!haschild) {
			return true;
		}
		return vs;
	}

}
