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 sys
# try:
# 	sys.stdin = open('input.txt', 'r') 
# 	sys.stdout = open('output.txt', 'w')
# except:
# 	pass

input = sys.stdin.readline
def DFS(i):
	visited = {i:True}
	stack = [(i,0)]
	while len(stack)!=0:
		tail,depth = stack.pop(-1)
		flag = True
		for each in neigh[tail]:
			if each not in visited:
				visited[each] = True
				flag = False
				stack.append((each,depth+1))
		if flag:
			leafDepth.append(depth)


n = int(input())

neigh = [[] for i in range(n)]
l = []
for i in range(n - 1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    neigh[a].append(b)
    neigh[b].append(a)
    l.append((a,b))
#Max
edges = set()
for a, b in l:
    if len(neigh[a]) == 1:
        a = -1
    if len(neigh[b]) == 1:
        b = -1
    if a > b:
        a, b = b, a
    edges.add((a,b))

MAX = len(edges)

#Min
leafDepth = []

DFS(0)
if (len(neigh[0])==1):
	MIN = 1 if all([True if i%2==0 else False for i in leafDepth]) else 3
else:
	MIN = 1 if all([True if i%2==0 else False for i in leafDepth]) or all([True if i%2==1 else False for i in leafDepth]) else 3
        
#Out
print(MIN, MAX)