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.lang.*;
import java.io.*;
import java.math.*;
//BigInteger A;
//A= BigInteger.valueOf(54);
//ArrayList<Integer> a=new ArrayList<>();
//TreeSet<Integer> ts=new TreeSet<>();
//HashMap<Integer,Integer> hm=new HashMap<>();
//PriorityQueue<Integer> pq=new PriorityQueue<>();
public final class Practice
{
    public static void main(String[]args)throws IOException
    {
        int K=(int)Math.pow(10,9)+7;
        FastReader ob=new FastReader();
        //int t=ob.nextInt();
        //while(t-->0)
        {
            int n=ob.nextInt();
            ArrayList<ArrayList<Integer>> adj=new ArrayList<>();
            for(int i=0;i<n;i++)
            adj.add(new ArrayList<>());
            for(int i=0;i<n-1;i++)
            {
                int u=ob.nextInt();
                int v=ob.nextInt();
                --u;--v;
                adj.get(u).add(v);
                adj.get(v).add(u);
            }
            TreeSet<Integer> ts=new TreeSet<>();int leafcount=0,leaf=-1;
            for(int i=0;i<n;i++)
            {
                if(adj.get(i).size()==1)
                {
                    leaf=i;
                    ++leafcount;
                    ts.add(adj.get(i).get(0));
                }
            }
            int max=(n-1)-(leafcount-ts.size());
            Queue<Integer> q=new LinkedList<>();
            q.add(leaf);
            int level=0;
            boolean vis[]=new boolean[n];
            vis[leaf]=true;
            int min=1;
            while(!q.isEmpty())
            {
                int size=q.size();
                for(int i=0;i<size;i++)
                {
                    int x=q.poll();
                    if(adj.get(x).size()==1)
                    {
                        if(level%2!=0)
                        min=Math.max(min,3);
                        else
                        min=Math.max(min,1);
                    }
                    for(int j=0;j<adj.get(x).size();j++)
                    {
                        if(!vis[adj.get(x).get(j)])
                        {
                            q.add(adj.get(x).get(j));
                            vis[adj.get(x).get(j)]=true;
                        }
                    }
                }
                //System.out.println(q);
                ++level;
            }
            System.out.println(min+" "+max);
        }
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;
    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e)  {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    public String nextLine()
    {
        String s="";
        try {
        s=br.readLine();
        } catch (IOException e)  {
                e.printStackTrace();
            }
        return s;
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}