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, stdout

def main():
    n = int(stdin.readline())
    g = [[] for _ in range(n+1)]
    k = [[] for _ in range(n+1)]
    already = [0] * (n+1)
    already1 = [0] * (n+1)
    even = [False] * (n+1)
    odd = [False] * (n+1)
    for _ in range(n-1):
        a,b = list(map(int, stdin.readline().split()))
        g[a].append(b)
        g[b].append(a)
        k[a].append(b)
        k[b].append(a)
    start = -1
    for i in range(1,n+1):
        if len(g[i]) > 1:
            start = i
            break
    stack = [start]
    mn = 1
    mx = n - 1
    while stack:
        x = stack[-1]
        if already[x] == 0:
            already[x] = -1
        still_deep = False
        if len(g[x]) > 0:
            y = g[x][-1]
            if already[y] == 0:
                stack.append(y)
                g[x].pop()
                still_deep = True
                continue
            else:
                g[x].pop()
                continue
        if still_deep is True:
            continue
        stack.pop()
        if already[x] == -1 and len(k[x]) == 1:
            already[x] = 1
            already1[stack[-1]] += 1
            odd[stack[-1]] = True
            continue
        if odd[x] == even[x]:
            mn = 3
        if already1[x] > 0:
            mx -= (already1[x] - 1)
        already[x] = 1
        if stack:
            y = stack[-1]
            if odd[x] is True:
                even[y] = True
            if even[x] is True:
                odd[y] = True
              
                   
    stdout.write(str(mn) + " " + str(mx))
    
main()