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.*;
public class A
{
      static int leaves=0;
      static HashSet<Integer> hs[];
      static HashSet<Integer> h;
      static int neighbourToLeaf=0;
      static int even=0,odd=0;
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            int n=Integer.parseInt(br.readLine());
            hs=new HashSet[n+1];
            h=new HashSet<Integer>();
            
            for(int i=1;i<=n;i++)
             hs[i]=new HashSet<Integer>();
            
            for(int i=0;i<n-1;i++)
            {
                  String s1[]=br.readLine().split(" ");
                  int u=Integer.parseInt(s1[0]);
                  int v=Integer.parseInt(s1[1]);
                  hs[v].add(u);
                  hs[u].add(v);
            }
            
            Iterator it=hs[1].iterator();
            h.add(1);
            dfs(1,0);
            
            int min=0,max=0;
            if(even>0 && odd>0)
             min=3;
            else
             min=1;
            
            //System.out.println(even+" "+odd);
            max=n-1-leaves+neighbourToLeaf;
            System.out.println(min+" "+max);
      }
      
      public static void dfs(int u,int d)
      {
            Iterator it=hs[u].iterator();
            if(hs[u].size()==1)
            { 
              leaves++; 
              if(d%2==0)
               even++;
              else
               odd++;
            }
            
            boolean bb1=false;
            while(it.hasNext())
            {
                  int v=(int)it.next();
                  if(!h.contains(v))
                  {
                    h.add(v);
                    dfs(v,d+1);
                  }
                  if(hs[v].size()==1)
                    bb1=true;
            }
            if(bb1)
              neighbourToLeaf++;
      }
}