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) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(in.readLine());
        ArrayList<Integer>[] adj = new ArrayList[n+1];
        for (int i = 0; i <= n; i++) adj[i] = new ArrayList<>();
        for (int i = 0; i < n-1; i++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            int u = Integer.parseInt(st.nextToken());
            int v = Integer.parseInt(st.nextToken());
            adj[u].add(v);
            adj[v].add(u);
        }

        boolean[] isPar = new boolean[n+1];
        boolean[] isLeaf = new boolean[n+1];
        int leaves = 0;
        for (int i = 1; i <= n; i++) {
            if (adj[i].size() == 1) {
                isLeaf[i] = true;
                leaves++;
                isPar[adj[i].get(0)] = true;
            }
        }
        int pars = 0;
        int source = 0;
        for (int i = 1; i <= n; i++) {
            if (isPar[i]) {
                pars++;
                source = i;
            }
        }
        int max = n-1-leaves+pars;
        int[] dist = new int[n+1];
        Queue<Integer> q = new LinkedList<>();
        boolean[] vis = new boolean[n+1];
        vis[source] = true;
        q.offer(source);
        while (!q.isEmpty()) {
            int u = q.poll();
            for (int v : adj[u]) {
                if (!vis[v]) {
                    dist[v] = dist[u] + 1;
                    vis[v] = true;
                    q.offer(v);
                }
            }
        }
        boolean odd = false;
        boolean even = false;
        int min = 1;
        for (int i = 1; i <= n; i++) {
            if (!isLeaf[i]) {
                continue;
            }
            if (dist[i] % 2 == 0) {
                even = true;
            } else {
                odd = true;
            }
        }
        if (even && odd) {
            min = 3;
        }
        out.println(min + " " + max);
        out.close();
    }

}
/*

 */
