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>
from collections import defaultdict,deque
from sys import *
input = stdin.readline

n = int(input())
graph = defaultdict(list)
deg = [0] * n
for _ in range(n-1):
    a,b = map(int,input().split())
    graph[a-1].append(b-1)
    graph[b-1].append(a-1)
    deg[a-1]+=1
    deg[b-1]+=1

childLeaves = defaultdict(int)
idx = 0
for i in range(n):
    if deg[i]==1:
        idx=i
        childLeaves[graph[i][0]]+=1

mAx = n-1 +(len(childLeaves))-sum(childLeaves.values())

dis = defaultdict(int)
dis[idx]=0
stack = deque()
stack.append(idx)
visited = [0]*n
visited[idx] = 1
mIn = 1

while stack:
    node = stack.pop()
    if not visited[node]:
        visited[node]=1
    for i in graph[node]:
        if not visited[i]:
            dis[i] = dis[node]+1
            stack.append(i)
            if deg[i]==1 and dis[i]%2:
                mIn=3

print(mIn,mAx)