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>
#!/usr/bin/env python3
import sys
input = sys.stdin.readline

n = int(input())
edge = [[] for _ in range(n)]
for _ in range(n-1):
    u, v = map(int, input().split())
    u -= 1; v -= 1
    edge[u].append(v)
    edge[v].append(u)

is_leaf = [False] * n
for i in range(n):
    if len(edge[i]) == 1:
        is_leaf[i] = True
    else:
        start = i

oe = [0, 0]
def dfs(d, p, v):
    queue = [(d, p, v)]
    while queue:
        d, p, v = queue.pop()
        if is_leaf[v]:
            oe[d%2] += 1
            continue
        for nv in edge[v]:
            if nv == p:
                continue
            queue.append((d+1, v, nv))
dfs(0, -1, start)
if 0 in oe:
    minima = 1
else:
    minima = 3

dim = 0
for i in range(n):
    cnt = 0
    for v in edge[i]:
        cnt += is_leaf[v]
    if cnt > 1:
        dim += cnt - 1 
    
print(minima, n - 1 - dim)