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>
# Fast IO (only use in integer input)

import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

n = int(input())

connectionList = []
for _ in range(n):
    connectionList.append([])
for _ in range(n-1):
    p,q = map(int,input().split())
    connectionList[p-1].append(q-1)
    connectionList[q-1].append(p-1)

leavesList = []
for i in range(n):
    if len(connectionList[i]) == 1:
        leavesList.append(i)

distance = [-1] * n
distance[0] = 0
dfsStack = [(0,0)]

while dfsStack:
    elem,distanceCur = dfsStack.pop()
    for v in connectionList[elem]:
        if distance[v] == -1:
            dfsStack.append((v,distanceCur + 1))
            distance[v] = distanceCur + 1

mainDist = distance[leavesList[0]]
minAns = 1
for elem in leavesList:
    if (distance[elem] - mainDist) % 2 != 0:
        minAns = 3
        break

oneUpLeaveList = []
for elem in leavesList:
    oneUpLeaveList.append(connectionList[elem][0])

oneUpLeaveList.sort()
uniqueElement = 0
for i in range(len(oneUpLeaveList)):
    if i == 0 or oneUpLeaveList[i-1] != oneUpLeaveList[i]:
        uniqueElement += 1

maxAns = n - 1 - (len(oneUpLeaveList) - uniqueElement)
print(str(minAns) + " " + str(maxAns))