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 io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

from collections import deque 

n = int(input())

neigh = [[] for i in range(n)]
for i in range(n-1):
    u,v = map(int,input().split())
    neigh[v-1].append(u-1)
    neigh[u-1].append(v-1)
    
for i in range(n):
    if len(neigh[i])==1: 
        root = i
        break

queue = deque()
visited = [False]*n
visited[root] = True
queue.append([root,0])


leafneigh = {}

maximum = n-1

odd = 0
even = 0


while queue:
    [index,d] = queue.popleft()
    if len(neigh[index])==1:
        if neigh[index][0] not in leafneigh:  
            leafneigh[neigh[index][0]]= d
            if d%2==0:  odd += 1
            else: even += 1
        else:  maximum -= 1



    for ele in neigh[index]:
        if visited[ele]: continue 
        visited[ele] = True
        queue.append([ele,d+1])

if min(odd,even)==0: minimum = 1
else: minimum = 3
        
#print(leafneigh)
#print(odd)
#print(even)

print(minimum,maximum)
