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>
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class B
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         LinkedList<Integer>[] edges = new LinkedList[N+1];
         for(int i=1; i <= N; i++)
            edges[i] = new LinkedList<Integer>();
         for(int i=0; i < N-1; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            edges[a].add(b);
            edges[b].add(a);
         }
         //find root
         int root = 1;
         for(int i=1; i <= N; i++)
            if(edges[i].size() > 1)
            {
               root = i;
               break;
            }
         //maximize
         boolean[] leaf = new boolean[N+1];
         for(int i=1; i <= N; i++)
            if(edges[i].size() == 1)
               leaf[i] = true;
         int dpval = dfs(root, edges, leaf, new boolean[N+1]);
         int lc = 0;
         for(int i=1; i <= N; i++)
            if(leaf[i])
               lc++;
         int max = N-1-(lc-dpval);
         //minimize
         int min = 1;
         int[] dist = new int[N+1];
         genDist(root, edges, dist, new boolean[N+1]);
         boolean[] par = new boolean[2];
         for(int i=1; i <= N; i++)
         {
            if(root == i || !leaf[i])
               continue;
            par[dist[i]%2] = true;
         }
         if(par[0] && par[1])
            min+=2;
         System.out.println(min+" "+max);
      }
      public static void genDist(int curr, LinkedList<Integer>[] edges, int[] dist, boolean[] seen)
      {
         seen[curr] = true;
         for(int next: edges[curr])
            if(!seen[next])
            {
               dist[next] = dist[curr]+1;
               genDist(next, edges, dist, seen);
            }
      }
      public static int dfs(int curr, LinkedList<Integer>[] edges, boolean[] leaf, boolean[] seen)
      {
         seen[curr] = true;
         int val = 0;
         int res = 0;
         for(int next: edges[curr])
            if(!seen[next])
            {
               if(leaf[next])
                  val = 1;
               else
                  res += dfs(next, edges, leaf, seen);
            }
         res += val;
         return res;
      }
   }