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
from collections import deque
input = sys.stdin.readline

n = int(input())
info = [list(map(int, input().split())) for i in range(n - 1)]

tree = [[] for i in range(n)]
for a, b in info:
    a -= 1
    b -= 1
    tree[a].append(b)
    tree[b].append(a)
    
max_ = n - 1
for v in range(n):
    cnt = 0
    for nxt_v in tree[v]:
        if len(tree[nxt_v]) == 1:
            cnt += 1
    max_ -= max(cnt - 1, 0)
    
    
visited = [-1] * n
q = deque([0])
visited[0] = 0
while q:
    v = q.popleft()
    for nxt_v in tree[v]:
        if visited[nxt_v] != -1:
            continue
        visited[nxt_v] = visited[v] + 1
        q.append(nxt_v)


min_ = 3
for i in range(n):
    if len(tree[i]) == 1:
        if visited[i] % 2 == 0:
            break
else:
    min_ = 1

for i in range(n):
    if len(tree[i]) == 1:
        if (visited[i] + 1) % 2 == 0:
            break
else:
    min_ = 1
print(min_, max_)