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 collections import defaultdict, deque
import os


def get_neighbors(edges):
    neighbors = defaultdict(set, {})
    for v1, v2 in edges:
        if v1 != v2:
            neighbors[v1].add(v2)
            neighbors[v2].add(v1)

    return dict(neighbors)


def is_all_even(adj, root):
    seen = set()
    fringe = {root}
    dist = 1
    while fringe:
        new_fringe = set()
        for node in fringe:
            seen.add(node)
            for n in adj[node]:
                if len(adj[n]) == 1 and dist % 2 == 1:
                    return False
                elif n not in seen:
                    new_fringe.add(n)

        fringe = new_fringe
        dist += 1

    return True


def solve(edges):
    adj = get_neighbors(edges)
    leaves = [i for i, j in adj.items() if len(j) == 1]
    m = 1 if is_all_even(adj, leaves[0]) else 3
    parents = set(k for l in leaves for k in adj[l])
    M = len(adj) - (len(leaves) - len(parents)) - 1
    return m, M


def pp(input):
    n = int(input())
    edges = [tuple(map(int, input().split(" "))) for _ in range(n - 1)]
    print(" ".join(map(str, solve(edges))))


if "paalto" in os.getcwd():
    from string_source import string_source

    pp(
        string_source(
            """6
1 3
2 3
3 4
4 5
5 6"""
        )
    )

    pp(
        string_source(
            """6
1 3
2 3
3 4
4 5
4 6"""
        )
    )

    pp(
        string_source(
            """7
1 2
2 7
3 4
4 7
5 6
6 7"""
        )
    )
else:
    pp(input)
