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.*;

public class Solution {
    static ArrayList<Integer> tree[];
    static HashSet<Integer> leaves=new HashSet<>();

    public static void main(String[] args) {
        Scanner input=new Scanner(System.in);
        int n=input.nextInt();
        tree=new ArrayList[n+1];
        for (int i = 0; i <=n ; i++) {
            tree[i]=new ArrayList<>();
        }
        int root=1;
        for (int i = 0; i <n-1 ; i++) {
            int x=input.nextInt();
            int y=input.nextInt();
            tree[x].add(y);
            tree[y].add(x);
            if (tree[x].size()>1){
                root=x;
            }
            if (tree[y].size()>1){
                root=y;
            }
        }
        int le=0;
        for (int i = 1; i <=n ; i++) {
            if (tree[i].size()==1){
                le=i;
                leaves.add(i);

            }

        }


        dfsminfind(le,0);
        dfsmaxfind(root);
        if (odddist){
            System.out.print(3+" ");
        }else {
            System.out.print(1+" ");
        }
        System.out.println(n-1 -leaves.size()+max);







    }
    static  int max=0;
    static HashSet<Integer> vismax=new HashSet<>();
    private static void dfsmaxfind(int root) {
            vismax.add(root);
        boolean found=false;
        for (int node:tree[root]) {
            if (vismax.contains(node)){
                continue;
            }
            if (tree[node].size()==1){
                if (!found){
                    found=true;
                    max++;
                }
                continue;
            }
            dfsmaxfind(node);

        }
    }

    static boolean odddist=false;
     static HashSet<Integer> vis=new HashSet<>();

    private static void dfsminfind(int leaf,int dist) {
        if (tree[leaf].size()==1 && dist!=0){
            if (dist%2!=0){
                odddist=true;
            }
            return;
        }
        vis.add(leaf);

        for (int node:tree[leaf]) {
            if (vis.contains(node)){
                continue;
            }
            dfsminfind(node,dist+1);
            if (odddist){
                return;
            }

        }
    }


}