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

n = int(sys.stdin.readline().strip())
N = [[] for i in range (0, n)]
C = [-1 for i in range (0, n)]
for i in range (0, n-1):
    a, b = list(map(int,sys.stdin.readline().strip().split()))
    a, b = a-1, b-1
    N[a].append(b)
    N[b].append(a)
C[0] = 0
L0 = [0]
L1 = []
while len(L0) + len(L1) > 0:
    for i in L0:
        for j in N[i]:
            if C[j] == -1:
                C[j] = 1
                L1.append(j)
    L0 = []
    for i in L1:
        for j in N[i]:
            if C[j] == -1:
                C[j] = 0
                L0.append(j)
    L1 = []
f = 1
F = n-1
L = []
M = []
for i in range (0, n):
    if len(N[i]) == 1:
        L.append(N[i])
        M.append(i)
L.sort()
for i in range (0, len(L)-1):
    if L[i] == L[i+1]:
        F = F - 1
    if C[M[i]] != C[M[i+1]]:
        f = 3
print(f, F)