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
from array import array  # noqa: F401
import typing as Tp  # noqa: F401


def input():
    return sys.stdin.buffer.readline().decode('utf-8')


def main():
    n = int(input())
    adj = [[] for _ in range(n)]
    deg = [0] * n
    for u, v in (map(int, input().split()) for _ in range(n - 1)):
        adj[u - 1].append(v - 1)
        adj[v - 1].append(u - 1)
        deg[u - 1] += 1
        deg[v - 1] += 1

    start = [i for i in range(n) if deg[i] == 1][0]
    stack = [(start, -1, 0)]
    ans_min, ans_max = 1, -1

    while stack:
        v, par, dist = stack.pop()
        ans_max += 1
        dist += 1

        has_leaf = 0

        for dest in adj[v]:
            if dest == par:
                continue
            if deg[dest] == 1:
                # dest is leaf
                has_leaf = 1
                if dist & 1:
                    ans_min = 3
                continue

            stack.append((dest, v, dist))

        if dist != 2:
            ans_max += has_leaf

    print(ans_min, ans_max)


if __name__ == '__main__':
    main()
