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>
mod = 1000000007
eps = 10**-9


def main():
    import sys
    input = sys.stdin.buffer.readline

    N = int(input())
    adj = [[] for _ in range(N+1)]
    for _ in range(N-1):
        a, b = map(int, input().split())
        adj[a].append(b)
        adj[b].append(a)

    from collections import deque
    que = deque()
    que.append(1)
    seen = [-1] * (N+1)
    seen[1] = 0
    par = [0] * (N+1)
    child = [[] for _ in range(N+1)]
    seq = []
    while que:
        v = que.popleft()
        seq.append(v)
        for u in adj[v]:
            if seen[u] == -1:
                seen[u] = seen[v] + 1
                par[u] = v
                child[v].append(u)
                que.append(u)
    seq.reverse()

    ok = 1
    flg = -1
    is_leaf = [0] * (N+1)
    for v in range(1, N+1):
        if len(adj[v]) == 1:
            if flg == -1:
                flg = seen[v] & 1
            else:
                if flg != seen[v] & 1:
                    ok = 0
            is_leaf[v] = 1

    if ok:
        m = 1
    else:
        m = 3
    M = N-1
    for v in range(1, N+1):
        cnt = 0
        for u in adj[v]:
            if is_leaf[u]:
                cnt += 1
        if cnt:
            M -= cnt-1
    print(m, M)


if __name__ == '__main__':
    main()
