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>
# CodeForces
# problem2
from collections import deque
from collections import defaultdict
N = int(input())
G = [[] for i
     in range(N+1)]
for i in range(N-1):
    a, b = map(int, input().split())
    G[a].append(b)
    G[b].append(a)

#maxi is oubious
maxi = N-1
S = -1
Leaves = []
D = defaultdict(int)
for i in range(1, N+1):
    if len(G[i]) == 1:
        S = i
        D[G[i][0]] += 1
        Leaves.append(i)

for value in D.values():
    maxi -= value-1

# mini case

q = deque()
visited = [False for i in range(N+1)]
dis = [0 for i in range(N+1)]
visited[S] = True
q.append(S)
while q:
    x = q.popleft()
    for node in G[x]:
        if visited[node] == False:
            dis[node] = dis[x]+1
            visited[node] = True
            q.append(node)
if all(dis[S] % 2 == dis[node] % 2 for node in Leaves):
    mini = 1
else:
    mini = 3
print(mini, maxi)