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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;

public class Main {

    public void solve(InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int[] leaves_count = new int[n]; // 1,2,3
        Map<Integer, Set<Integer>> g = new HashMap<>();

        isMinOne = true;

        for (int i = 1; i < n; i++) {
            int a = in.nextInt()-1;
            int b = in.nextInt()-1;
            Set<Integer> as = g.getOrDefault(a, new HashSet<>());
            Set<Integer> bs = g.getOrDefault(b, new HashSet<>());
            as.add(b);
            bs.add(a);
            g.put(a, as);
            g.put(b, bs);
        }

        int root = -1;
        for (int i = 0; i < n; i++) {
            if (g.get(i).size() == 1) {
                root = i;
                leaves_count[i] = -1;
            } else {
                leaves_count[i] = 0;
            }
        }

        for (int i = 0; i < n; i++) {
            if (leaves_count[i] == -1) {
                for (int r :g.get(i)) {
                    leaves_count[r]++;
                }
            }
        }

        //dfs from leaf;
        dfs(g,leaves_count, root, n);
        int min;
        if (isMinOne) min = 1;
        else min = 3;

        int max = n-1;
        for (int i: leaves_count) {
            if (i > 0) {
                max = max - (i-1);
            }
        }
        out.println(min + " " + max);
    }

    boolean isMinOne;
    public void dfs(Map<Integer, Set<Integer>> g, int[] vertices, int root, int n) {
        boolean[] visited = new boolean[n];
        visited[root] = true;
        dfs(g, vertices, root, 10, visited);
    }

    private void dfs(Map<Integer, Set<Integer>> g, int[] vertices, int root, int val, boolean[] visited) {
        for (int ch: g.get(root)) {
            if (visited[ch]) continue;
            if (vertices[ch] == -1) {
                if (val != 0) {
                    isMinOne = false;
                }
            }
            visited[ch] = true;
            dfs(g, vertices, ch, val^10, visited);
        }
    }

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Main obj = new Main();
        obj.solve(in, out);
        out.close();
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public float nextFloat() {
            return Float.parseFloat(next());
        }

    }
}
