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>
//package codeforces;

import java.io.*;
import java.util.*;

public class B1338 {

    public static void main(String[] args) throws IOException {
        BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(sc.readLine());
        Set<Integer>[] graph = new Set[n + 1];
        List<Integer> leaves = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            graph[i] = new HashSet<>();
        }
        for (int i = 1; i < n; i++) {
            String[] line = sc.readLine().split(" ");
            int from = Integer.parseInt(line[0]);
            int to = Integer.parseInt(line[1]);
            graph[from].add(to);
            graph[to].add(from);
        }
        if (n == 3) {
            System.out.println("1 1");
        } else {
            for (int i = 1; i <= n; i++) {
                if (graph[i].size() == 1) {
                    leaves.add(i);
                }
            }

            int[] leaveCount = new int[n + 1];
            for (int leaf : leaves) {
                for (int next : graph[leaf]) {
                    leaveCount[next]++;
                }
            }

            int[] distFromLeaf = bfs(graph, n, leaves.get(0));
            boolean hasOddPath = false;
            for (int leaf : leaves) {
                if (distFromLeaf[leaf] % 2 == 1) {
                    hasOddPath = true;
                    break;
                }
            }

            int max = n - 1;
            for (int i = 1; i <= n; i++) {
                if (leaveCount[i] > 1) {
                    max -= leaveCount[i];
                    max++;
                }
            }

            System.out.println((hasOddPath ? "3" : "1") + " " + max);
        }
    }

    private static int[] bfs(Set<Integer>[] graph, int n, int first) {
        List<Integer> currentLevel = new ArrayList<>();
        int currentDistance = 0;
        int[] dist = new int[n + 1];
        int maxDist = 0;
        int maxDistIndex = -1;
        for (int i = 1; i <= n; i++) {
            dist[i] = Integer.MAX_VALUE;
        }
        currentLevel.add(first);
        while (!currentLevel.isEmpty()) {
            maxDist = Math.max(maxDist, currentDistance);
            maxDistIndex = currentLevel.get(0);
            List<Integer> nextLevel = new ArrayList<>();
            for (int c : currentLevel) {
                dist[c] = currentDistance;
                for (int next : graph[c]) {
                    if (dist[next] == Integer.MAX_VALUE) {
                        nextLevel.add(next);
                    }
                }
            }
            currentLevel = nextLevel;
            currentDistance++;
        }
        return dist;
    }
}
