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 javaapplication5;

import java.io.*;
import java.util.*;
 
public class utkarsh {
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);
    
    ArrayList <Integer> a[];
    int level[];
    
    void dfs(int s, int pt) {
        for(int x : a[s]) {
            if(x == pt) continue;
            level[x] = level[s] + 1;
            dfs(x, s);
        }
    }
    
    void solve() {
        int n = ni();
        a = new ArrayList[n];
        for(int i = 0; i < n; i++)  a[i] = new ArrayList<>();
        for(int i = 0; i < n-1; i++) {
            int u = ni() - 1;
            int v = ni() - 1;
            a[u].add(v);    a[v].add(u);
        }
        int root = -1;
        for(int i = 0; i < n; i++) {
            if(a[i].size() == 1) {
                root = i;   break;
            }
        }
        level = new int[n];
        dfs(root, -1);
        int minf = 1;
        for(int i = 0; i < n; i++) {
            if(i == root)   continue;
            if(a[i].size() == 1 && level[i] % 2 == 1 && level[i] > 1) {
                minf = 3;   break;
            }
        }
        int maxf = n-1;
        for(int i = 0; i < n; i++) {
            int cnt = 0;
            for(int x : a[i]) {
                if(a[x].size() == 1)    cnt++;
            }
            if(cnt > 0) maxf -= (cnt - 1);
        }
        out.println(minf +" "+ maxf);
    }
    
    long mp(long b, long e) {
        long r = 1;
        while(e > 0) {
            if( (e&1) == 1 )    r = (r * b) % mod;
            b = (b * b) % mod;
            e >>= 1;
        }
        return r;
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }
    
    void run() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    public static void main(String[] args) {
        new utkarsh().run();
    }
}
