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
input = sys.stdin.buffer.readline
from collections import deque
def prog():
    n = int(input())
    adj_list = [[] for i in range(n+1)]
    for i in range(n-1):
        a,b = map(int,input().split())
        adj_list[a].append(b)
        adj_list[b].append(a)
    leaves = set()
    for i in range(1,n+1):
        if len(adj_list[i]) == 1:
            leaves.add(i)
    for i in leaves:
        root = i
        break
    q = deque([root])
    dists = [0 for i in range(n+1)]
    visited = set()
    while q:
        curr = q.popleft()
        for neighbor in adj_list[curr]:
            if neighbor not in visited:
                dists[neighbor] = dists[curr] + 1
                q.append(neighbor)
                visited.add(neighbor)
    odd = False
    for leave in leaves:
        if dists[leave] % 2 == 1:
            odd = True
            break
    if odd:
        minimum = 3
    else:
        minimum = 1
    distinct_leaves = 0
    for i in range(1,n+1):
        for neighbor in adj_list[i]:
            if neighbor in leaves:
                distinct_leaves += 1
                break
    print(minimum, n-1 - len(leaves) + distinct_leaves)
prog()
        
