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 sys import stdin

input = stdin.readline


class N:
    def __init__(self, v) -> None:
        self.v = v
        self.c = set()
        self.l = False


if __name__ == '__main__':
    n = int(input())

    arr = [N(i) for i in range(n)]
    es = set()

    for _ in range(n - 1):
        x, y = map(int, input().split())
        x -= 1
        y -= 1
        x, y = min(x, y), max(x, y)
        es.add((x, y))
        arr[x].c.add(arr[y])
        arr[y].c.add(arr[x])

    ps = set()

    for x in arr:
        if len(x.c) == 1:
            x.l = True
            ps.add(list(x.c)[0])

    for p in ps:
        kept = False
        rs = []
        for c in p.c:
            if c.l:
                if kept:
                    x = p.v
                    y = c.v
                    x, y = min(x, y), max(x, y)
                    es.discard((x, y))
                    rs.append(c)
                else:
                    kept = True
        for c in rs:
            p.c.discard(c)

    root = ps.pop()
    stk = [(root, 0)]
    viz = set()
    ls = []

    while stk:
        root, l = stk.pop()
        viz.add(root)
        for c in root.c:
            if c not in viz:
                if c.l:
                    ls.append(l + 1)
                else:
                    stk.append((c, l + 1))

    if len(es) > 1:
        print(1 + 2 * int(len(set([l % 2 for l in ls])) == 2), len(es))
    else:
        print(1, 1)
