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
import math
from collections import deque


def solve():
    #f = open("b.txt")
    f = sys.stdin
    n = int(f.readline())

    edges = [[] for i in range(n)]
    for i in range(n - 1):
        s = f.readline().strip()
        a, b = [int(x) for x in s.split(' ', 2)]
        edges[a - 1].append(b - 1)
        edges[b - 1].append(a - 1)

    root = 0
    for i in range(n):
        if len(edges[i]) > 1:
            root = i
            break

    #print(root)
    odd_dist = False
    even_dist = False
    minf = 1
    maxf = n - 1

    dist = [0 for i in range(n)]
    q = deque()
    q.append(root)
    while len(q) > 0:
        cur = q.popleft()
        child_leaves = 0
        for next in edges[cur]:
            if dist[next] == 0 and next != root:
                q.append(next)
                dist[next] = dist[cur] + 1
            if len(edges[next]) == 1:
                child_leaves += 1
                if dist[next] % 2 == 1:
                    odd_dist = True
                else:
                    even_dist = True
        if child_leaves > 1:
            maxf = maxf - child_leaves + 1

    if odd_dist and even_dist:
        minf = 3
    print(str(minf) + " " + str(maxf))


solve()