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 collections
from collections import defaultdict
import sys
input=sys.stdin.readline
n=int(input())
graph=defaultdict(list)
for i in range(n-1):
    a,b=map(int,input().split())
    graph[a].append(b)
    graph[b].append(a)
count=0
leavepar=[0]*(n+1)
leave=[]
for i in graph:
    if len(graph[i])==1:
        leave.append(i)
        leavepar[graph[i][0]]+=1
        if leavepar[graph[i][0]]>=2:
            count+=1
def bfs(graph, root):
    visited=set()
    dist=[0]*(n+1)
    queue=collections.deque([root])
    visited.add(root)
    while queue:
        vertex=queue.popleft()
        for neighbour in graph[vertex]:
            if neighbour not in visited:
                dist[neighbour]=dist[vertex]+1
                visited.add(neighbour)
                queue.append(neighbour)
    return(dist)
ret=bfs(graph,leave[0])
ans=1
for i in leave:
    if ret[i]%2==1:
        ans=3
        break
sys.stdout.write(str(ans)+' '+str(n-count-1)+'\n')
        
    
