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 deque
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(300000)

def dfs(root):
	stack=[root]
	while len(stack)!=0:
		element = stack.pop()
		for i in tree[element]:
			if num[i]==-1:
				stack.append(i)
				num[i]=1-num[element]


def bfs(root):
	queue=[root]
	queue=deque(queue)
	vis=[-1]*n
	while len(queue)!=0:
		element = queue.popleft()
		vis[element]=1
		for i in tree[element]:
			if vis[i]==-1:
				queue.append(i)
				child[element].append(i)
				vis[i]=1
			if len(tree[element])==1:
				leaf.append(element)


n=int(input())
tree=[[] for i in range(n)]
child=[[] for i in range(n)]
for i in range(n-1):
	a,b=list(map(int,input().split()))
	tree[a-1].append(b-1)
	tree[b-1].append(a-1)
leaf=[]
bfs(0)
num=[-1]*n
num[leaf[0]]=0
dfs(leaf[0])
minimum=1
for i in leaf:
	if num[i]!=num[leaf[0]]:
		minimum=3
		break
maximum=n-1
leaf = set(leaf)
for i in range(n):
	flag=0
	for j in tree[i]:
		if j in leaf:
			maximum-=1
			flag=1
	if flag:
		maximum+=1
print (minimum,maximum)
