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 random

def getVal(used):
    x = random.randint(1,10**40)
    while x in used:
        x = random.randint(1,10**40)

    return x

def solve(edges,n,leaf):
    stack = [leaf]
    dp = [0]*(n+1)
    dp2 = [0]*(n+1)
    one = True
    visited = set()
    used = set()
    while stack:
        curr = stack.pop()
        if curr not in visited:
            visited.add(curr)
            if len(edges[curr]) == 1 and dp2[curr] != 0:
                one = False
            for kid in edges[curr]:
                if kid not in visited:
                    dp2[kid] = dp2[curr]^1
                    stack.append(kid)
                    val = getVal(used)
                    if len(edges[kid]) == 1:
                        used.add(dp[curr])
                        dp[kid] = 0
                    else:
                        dp[kid] = dp[curr]^val
                        used.add(val)


    if one:
        min_val = 1
    else:
        min_val = 3
        
    print(min_val,len(used))

def main():
    n = int(input())
    edges = {}
    for i in range(1,n+1):
        edges[i] = []

    for i in range(n-1):
        a,b = map(int,input().split())
        edges[a].append(b)
        edges[b].append(a)


    for i in edges.keys():
        if len(edges[i]) == 1:
            leaf = i
            break

    solve(edges,n,leaf)


main()
