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 {
    private static ArrayList<Integer>[] adjacencyList;
    private static int dfs(int root, int parent, int depth) {
        boolean[] parity = new boolean[2];
        for(int i: adjacencyList[root]) {
            if(i != parent) {
                int temp = dfs(i, root, depth+1);
                if(temp == 2) {
                    return 2;
                }
                parity[temp] = true;
            }
        }
        if(parity[0] && parity[1]) {
            return 2;
        }
        if(!parity[0] && !parity[1]) {
            parity[depth%2] = true;
        }
        return parity[0] ? 0 : 1;
    }
    public static void main(String[] args) throws IOException{
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n = Integer.parseInt(f.readLine());
        adjacencyList = new ArrayList[n+1];
        for(int i = 1; i <= n; i++) {
            adjacencyList[i] = new ArrayList<>();
        }
        int[] degree = new int[n+1];
        for(int i = 0; i < n-1; i++) {
            StringTokenizer st = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            adjacencyList[a].add(b);
            adjacencyList[b].add(a);
            degree[a]++;
            degree[b]++;
        }
        int temp = dfs(1, 0, 0);
        int min = temp == 2 || (degree[1] == 1 && temp > 0) ? 3 : 1;
        int max = n-1;
        for(int i = 1; i <= n; i++) {
            if(degree[i] == 1) {
                max--;
            } else {
                for(int j: adjacencyList[i]) {
                    if(degree[j] == 1) {
                        max++;
                        break;
                    }
                }
            }
        }
        out.println(min + " " + max);
        f.close();
        out.close();
    }
}