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.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
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;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            List<Integer>[] adj = new List[n];
            for (int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }
            int[] deg = new int[n];
            for (int i = 0; i < n - 1; i++) {
                int a = in.nextInt() - 1;
                int b = in.nextInt() - 1;
                adj[a].add(b);
                adj[b].add(a);
                ++deg[a];
                ++deg[b];
            }
            int root = -1;
            for (int i = 0; i < n; i++) {
                if (deg[i] > 1) {
                    root = i;
                }
            }
            if (root < 0) {
                throw new AssertionError();
            }

            Solution ans = dfs(adj, root, -1);
            out.println(ans.min + " " + ans.max);
        }

        private Solution dfs(List<Integer>[] adj, int v, int p) {
            Solution s = new Solution();
            s.isLeaf = true;
            boolean firstLeaf = true;
            s.min = 1;
            for (int u : adj[v]) {
                if (u == p) {
                    continue;
                }
                s.isLeaf = false;
                Solution t = dfs(adj, u, v);
                s.min = Math.max(s.min, t.min);
                if (t.hasEven) {
                    s.hasOdd = true;
                } else {
                    s.hasEven = true;
                }
                if (t.isLeaf) {
                    if (firstLeaf) {
                        firstLeaf = false;
                        ++s.max;
                    }
                } else {
                    s.max += t.max + 1;
                }
            }
            if (s.isLeaf) {
                s.hasOdd = true;
            }
            if (s.hasEven && s.hasOdd) {
                s.min = 3;
            }
            return s;
        }

        class Solution {
            long min;
            long max;
            boolean isLeaf;
            boolean hasEven;
            boolean hasOdd;

        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

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

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

    }
}

