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
input = sys.stdin.readline

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 = []
visited = [False] * n
from collections import deque
q = deque()
q.append((0,0))
while q:
    nex, d = q.popleft()
    if not visited[nex]:
        visited[nex] = True

        if len(neigh[nex]) == 1:
            leafDepth.append(d)
        for v in neigh[nex]:
            q.append((v,d+1))

MIN = 1
corr = leafDepth.pop() % 2
for d in leafDepth:
    if d % 2 != corr:
        MIN = 3 
        


#Out
print(MIN, MAX)
