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>
from __future__ import division,print_function
import sys
if True:
    le = sys.__stdin__.read().split("\n")
    le.pop()
    le = le[::-1]
    n = int(le.pop())
    ar = [[] for k in range(n)]
for q in le:
    a,b = map(int,q.split())
    a-=1
    b-=1
    ar[a].append(b)
    ar[b].append(a)
le = []
p = [-1]*n
fe = [v for v in range(n) if len(ar[v])==1]
r = min([v for v in range(n) if len(ar[v])!=1])
de = [-1]*n
de[r] = 0
pi = [r]
while pi:
    v=pi.pop()
    for w in ar[v]:
        if p[v] != w:
            p[w] = v
            de[w] = de[v]+1
            pi.append(w)
zor = len({de[v]%2 for v in fe})==2
s = 0
for l in ar:
    a = 0
    for v in l:
        a += len(ar[v])==1
    s += max(0,a-1)
print(1+2*zor,n-1-s)#sum(max(0,sum(len(ar[v])==1 for v in l)-1) for l in ar))
