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>
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
    static ArrayList<ArrayList<Integer>> arr = new ArrayList<ArrayList<Integer>>();
    static boolean vis[];
    static boolean flag = false;
	public static void main (String[] args) throws java.lang.Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb = new StringBuffer();
		int n = Integer.parseInt(br.readLine());
		for(int i = 0; i <= n; i++){
		    arr.add(new ArrayList<Integer>());
		}
		for(int i = 0; i < n - 1; i++){
		    String s[] = (br.readLine()).split(" ");
		    int a = Integer.parseInt(s[0]);
		    int b = Integer.parseInt(s[1]);
		    arr.get(a).add(b);
		    arr.get(b).add(a);
		}
		int min = 1, max = n - 1;
		int start = 0;
		for(int i = 1; i <= n; i++){
		    if(arr.get(i).size() == 1) start = i;
		    else{
		        int count = 0;
		        for(int j = 0; j < arr.get(i).size(); j++){
		            if(arr.get(arr.get(i).get(j)).size() == 1) count++;
		        }
		        if(count > 1) max -= (count - 1);
		    }
		}
		vis = new boolean[n + 1];
		dfs(start, -1, 0);
		if(flag) min = 3;
		System.out.println(min+" "+max);
	}
	public static void dfs(int node, int par, int val){
	    vis[node] = true;
	    if(arr.get(node).size() == 1 && val % 2 == 1) flag = true;
	    for(int i = 0; i < arr.get(node).size(); i++){
	        int t = arr.get(node).get(i);
	        if(!vis[t]) dfs(t, node, val + 1);
	    }
	}
}
