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 
n1 = input()
n = int(n1)
if n == 2: 
    print(1, 1)
    exit()
edges = [set() for _ in range(n)]
for i in range(n - 1):
    x = input()
    a, b = x.split(' ')
    a, b = int(a), int(b)
    edges[a-1].add(b-1)
    edges[b-1].add(a-1)
ocnt = collections.deque()
for i in range(n):
    if len(edges[i]) == 1:
        ocnt.append(i)
nl = set()

for i in ocnt:
    for j in edges[i]:
        nl.add(j)

x = ocnt[0]
que, l = collections.deque([(-1, x)]), 1
lo = 1
while que:
    size = len(que)
    for _ in range(size):
        p, top = que.popleft()
        if len(edges[top]) == 1 and l % 2 == 0:
            lo = 3
            break
        for j in edges[top]:
            if j != p:
                que.append((top, j))
    l += 1

hi = n - len(ocnt) + len(nl) - 1
print(lo, hi)

