1339_D. Edge Weight Assignment
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>
```python
import sys
lines = sys.stdin.readlines()
n = int(lines[0].strip())
edges = {}
for i in range(1,n):
    (u, v) = map(int, lines[i].strip().split(" "))
    if u not in edges: edges[u] = []
    if v not in edges: edges[v] = []
    edges[u].append(v)
    edges[v].append(u)

leaves = []
parLeave = {}
for u in edges.keys():
    if len(edges[u]) == 1: 
        leaves.append(u)
        parLeave[u] = edges[u][0]

maxF = n - (len(leaves) - len(set(parLeave.values()))) - 1

parity = {leaves[0]: 0}
stack = [leaves[0]]
seen = set(stack)
while stack:
    node = stack.pop()
    for adj in edges[node]:
        if adj not in seen:
            parity[adj] = 1- parity[node]
            stack.append(adj)
            seen.add(adj)
parityOfLeaves = list(map(lambda x: parity[x], leaves))
if min(parityOfLeaves) == max(parityOfLeaves): minF = 1
else: minF = 3
print("{} {}".format(minF, maxF))
```