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.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;
import static java.lang.Long.bitCount;

public class Main {

    static class Graph{
        ArrayList<Integer> []G;
        int []dp;
        int count_leaves = 0;
        boolean []visit;
        Graph(int N){
            dp = new int[N+1];
            Arrays.fill(dp, 0);
            visit = new boolean[N+1];
            Arrays.fill(visit, false);
            G = new ArrayList [N+1];
            for(int i=1; i<=N; ++i){
                G[i] = new ArrayList<Integer> ();
            }
        }

        void add(int i, int j){
            G[i].add(j);
            G[j].add(i);
        }

        int get_leaf(){
            int ans = -1;
            for(int i=1; i<G.length; ++i){
                if(ans == -1 && G[i].size() == 1){
                    ans = i;
                }
                if(G[i].size()==1) ++count_leaves;
            }
            return ans;
        }

        void dfs(int n, int counter){
            if(visit[n]) return;
            visit[n] = true;

            dp[n] = counter;
            ++counter;
            for(int x: G[n]){
                dfs(x, counter);
            }
        }

        void display_dist(){
            for(int x: dp) System.out.print(x+" ");
            System.out.println();
        }

        int same_parity(){
            int counter=0;
            for(int i=1; i<G.length; ++i){
                if(G[i].size()==1 && dp[i]%2==0) ++counter;
            }
            if(counter==count_leaves) return 1;
            else return 3;
        }

        int max_xor(){
            int counter=0;

            for(int i=1; i<G.length; ++i){
                int cc = 0;
                for(int x: G[i]){
                    if(G[x].size() == 1) ++cc;
                }
                --cc;
                counter += Math.max(0, cc);
            }

            return G.length-2-counter;
        }
    }

    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int N = in.nextInt();
        Graph G = new Graph(N);

        for(int i=1; i<=N-1; ++i){
            int u, v;
            u = in.nextInt();
            v = in.nextInt();
            G.add(u, v);
        }

        int start_node = G.get_leaf();
        //System.out.println(start_node);
        G.dfs(start_node, 0);

       // G.display_dist();
       // System.out.println();

        System.out.print(G.same_parity()+" ");
        System.out.println(G.max_xor());
    }

    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }

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