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 Main
{
    //static int even,odd;
    public static void main(String args[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        int n=Integer.parseInt(br.readLine());
        //even=odd=0;
        ArrayList<Integer> leaves=new ArrayList<>();
        ArrayList<Integer> edge[]=new ArrayList[n+1];
        TreeMap<Integer,Integer> tmap=new TreeMap<>();
        int deg[]=new int[n+1];
        int depth[]=new int[n+1];
        boolean vis[]=new boolean[n+1];
        for(int i=1;i<=n;i++)
        edge[i]=new ArrayList<>();
        for(int i=0;i<n-1;i++)
        {
            String str[]=br.readLine().split(" ");
            int u=Integer.parseInt(str[0]);
            int v=Integer.parseInt(str[1]);
            deg[u]++;
            deg[v]++;
            edge[u].add(v);
            edge[v].add(u);
        }
        for(int i=1;i<=n;i++)
        {
            if(deg[i]==1)
            leaves.add(i);
        }
        int src=leaves.get(0);
        vis[src]=true;
        Queue<Integer> queue=new LinkedList<>();
        queue.add(src);
        while(!queue.isEmpty())
        {
            int node=queue.poll();
            for(int i=0;i<edge[node].size();i++)
            {
                if(!vis[edge[node].get(i)])
                {
                    vis[edge[node].get(i)]=true;
                    queue.add(edge[node].get(i));
                    depth[edge[node].get(i)]=depth[node]+1;
                }
            }
        }
        int m=0;
        for(int i=1;i<=n;i++)
        {
            if(Collections.binarySearch(leaves,i)<0)
            {
                for(int j=0;j<edge[i].size();j++)
                {
                    int tp=edge[i].get(j);
                    if(Collections.binarySearch(leaves,tp)>=0)
                    {
                        m++;
                        break;
                    }
                }
            }
        }
        boolean isOdd=false;
        for(int i=0;i<leaves.size();i++)
        if(depth[leaves.get(i)]%2!=0)
        isOdd=true;
        if(isOdd)
        pw.print("3 ");
        else
        pw.print("1 ");
        pw.println((n-1)-leaves.size()+m);
        pw.flush();
        pw.close();
    }
}