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.*;
import java.util.*;
public class Main {
    public static void main(String args[])
    {
        FastReader input=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int T=1;
        while(T-->0)
        {
            int n=input.nextInt();
            ArrayList<Integer> adj[]=new ArrayList[n+1];
            ArrayList<Integer> p[]=new ArrayList[n+1];
            Vertex V[]=new Vertex[n+1];
            for(int i=1;i<=n;i++)
            {
                adj[i]=new ArrayList<>();
                V[i]=new Vertex(i);
                p[i]=new ArrayList<>();
            }
            for(int i=0;i<n-1;i++)
            {
                int u=input.nextInt();
                int v=input.nextInt();
                adj[u].add(v);
                adj[v].add(u);
            }
            int s=1;
            for(int i=1;i<=n;i++)
            {
                if(adj[i].size()>1)
                {
                    s=i;
                    break;
                }
            }
            bfs(adj,V,s);
            int min=1;
            int o=0,e=0;
            for(int i=1;i<=n;i++)
            {
                if(adj[i].size()==1)
                {
                    if(V[i].d%2==0)
                    {
                        o=1;
                    }
                    else
                    {
                        e=1;
                    }
                    int p1=V[i].parent.key;
                    p[p1].add(i);
                }
            }
            if(o==1 && e==1)
            {
                min=3;
            }
            int sum=0;
            int max=0;
            for(int i=1;i<=n;i++)
            {
                if(p[i].size()>=2)
                {
                    sum+=p[i].size();
                    max++;
                }
            }
            max+=n-1-sum;
            out.println(min+" "+max);
        }
        out.close();
    }
    public static void bfs(ArrayList<Integer> adj[],Vertex V[], int s)
    {
        V[s].color="gray";
        V[s].d=0;
        V[s].parent=null;
        Queue<Integer> q=new LinkedList<>();
        q.add(s);
        while(!q.isEmpty())
        {
            int k=q.poll();
            Vertex u=V[k];
            for(int i=0;i<adj[k].size();i++)
            {
                int key=adj[k].get(i);
                Vertex v=V[key];
                if(v.color=="white")
                {
                    q.add(key);
                    v.color="gray";
                    v.d=u.d+1;
                    v.parent=u;
                }
            }
            u.color="black";
        }
    }
    public static class Vertex
    {
        int key;
        int d,f;
        String color;
        Vertex parent;
        Vertex(int key)
        {
            this.key=key;
            this.d=0;
            this.f=0;
            this.color="white";
            this.parent=null;
        }
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong()
        {

            return Long.parseLong(next());
        }

        double nextDouble()
        {

            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}