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

n = int(input())

adj = [[] for _ in range(n)]

for _ in range(n-1):
    a, b = [int(x) for x in sys.stdin.readline().split()]
    a -= 1
    b -= 1
    adj[a].append(b)
    adj[b].append(a)

mx = n - 1

for i in range(n):
    num_adj_leaves = sum([1 for b in adj[i] if len(adj[b]) == 1])
    if num_adj_leaves:
        mx -= num_adj_leaves - 1

root = 0
parent = [None]*n
parent[0] = -1

stack = [0]

toposort = []

while len(stack):
    cur = stack.pop()
    toposort.append(cur)
    for nb in adj[cur]:
        if parent[nb] == None:
            parent[nb] = cur
            stack.append(nb)

toposort = list(reversed(toposort))
has_even_from_leaf = [False]*n
has_odd_from_leaf = [False]*n

for v in toposort:
    if len(adj[v]) == 1:
        has_even_from_leaf[v] = True

    if parent[v] != -1:
        has_even_from_leaf[parent[v]] |= has_odd_from_leaf[v]
        has_odd_from_leaf[parent[v]] |= has_even_from_leaf[v]

mn = 1
if any(has_even_from_leaf[v] and has_odd_from_leaf[v] for v in range(n)):
    mn += 2

print(mn, mx)
