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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        ArrayList<Integer>[] tree;
        boolean[] leaf;
        int[] cntLeaves;
        boolean odd = false;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int N = in.nextInt();
            tree = new ArrayList[N];
            leaf = new boolean[N];
            cntLeaves = new int[N];
            for (int i = 0; i < N; i++) {
                tree[i] = new ArrayList<>();
            }
            for (int i = 0; i < N - 1; i++) {
                int a = in.nextInt() - 1;
                int b = in.nextInt() - 1;
                tree[a].add(b);
                tree[b].add(a);
            }
            for (int i = 0; i < N; i++) {
                if (tree[i].size() == 1) {
                    leaf[i] = true;
                    for (int ch : tree[i]) {
                        cntLeaves[ch]++;
                    }
                }
            }
            int max = N - 1;
            for (int i = 0; i < N; i++) {
                if (cntLeaves[i] >= 2) {
                    max -= cntLeaves[i] - 1;
                }
            }
            for (int i = 0; i < N; i++) {
                if (!leaf[i]) {
                    dfs(i, i);
                    break;
                }
            }
            int min = 1;
            if (odd) {
                min = 3;
            }
            out.println(min);
            out.println(max);
        }

        int[] dfs(int n, int p) {
            if (leaf[n]) {
                return new int[]{1, 0};
            }
            int[] a = new int[]{0, 0};
            for (int ch : tree[n]) {
                if (ch != p) {
                    int[] b = dfs(ch, n);
                    a[0] += b[1];
                    a[1] += b[0];
                }
            }
            if (a[1] > 0 && a[0] > 0) {
                odd = true;
            }
            return a;
        }

    }

    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());
        }

    }
}

