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.*;

final public class B {

    static ArrayList<Integer>[] tree;
    static int[] dist;
    static HashSet<Integer> hs = new HashSet<>();
    static ArrayList<Integer> leaves = new ArrayList<>();

    static void dfs(int ver, int par, int level){
        dist[ver] = level;
        if(tree[ver].size() == 1 && tree[ver].get(0) == par){
            leaves.add(ver);
            hs.add(par);
        }
        for(int z: tree[ver]){
            if(z!=par){
                dfs(z, ver, level+1);
            }
        }
    }

    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int n = Integer.parseInt(br.readLine());
        tree = new ArrayList[n];
        dist = new int[n];
        for(int i = 0; i<n; i++){
            tree[i] = new ArrayList<>();
        }
        for(int i = 0; i<n-1; i++){
            String[] str = br.readLine().split(" ");
            int v = Integer.parseInt(str[0])-1, u = Integer.parseInt(str[1])-1;
            tree[v].add(u);
            tree[u].add(v);
        }
        int leaf = 0;
        for(int i = 0; i<n; i++){
            if(tree[i].size() == 1){
                leaf = i;
                hs.add(tree[i].get(0));
                break;
            }
        }
        leaves.add(leaf);
        dfs(leaf, -1, 0);
        int max = n-1 - leaves.size()+hs.size();
        boolean allEven = true;
        for(int z: leaves){
            if(dist[z]%2 != 0){
                allEven = false;
                break;
            }
        }
        int min = (allEven)?1:3;
        System.out.println(min+" "+max);
    }
}
