Solve in Python:
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 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 $n$ ($3 \le n \le 10^{5}$) — the number of vertices in given tree.

The $i$-th of the next $n-1$ lines contains two integers $a_{i}$ and $b_{i}$ ($1 \le a_{i} \lt b_{i} \le n$) — 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...
import sys
input = sys.stdin.readline

n = int(input())

neigh = [[] for i in range(n)]
l = []
for i in range(n - 1):
    a, b = list(map(int, input().split()))
    a -= 1
    b -= 1
    neigh[a].append(b)
    neigh[b].append(a)
    l.append((a,b))

#Max
edges = set()
for a, b in l:
    if len(neigh[a]) == 1:
        a = -1
    if len(neigh[b]) == 1:
        b = -1
    if a > b:
        a, b = b, a
    edges.add((a,b))

MAX = len(edges)

#Min
leafDepth = []
visited = [False] * n
from collections import deque
q = deque()
q.append((0,0))
while q:
    nex, d = q.popleft()
    if not visited[nex]:
        visited[nex] = True

        if len(neigh[nex]) == 1:
            leafDepth.append(d)
        for v in neigh[nex]:
            q.append((v,d+1))

MIN = 1
corr = leafDepth.pop() % 2
for d in leafDepth:
    if d % 2 != corr:
        MIN = 3 
        


#Out
print(MIN, MAX)