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 sys, os
input = sys.stdin.buffer.readline
n = int(input())
g = [[] for i in range(n+1)]
for i in range(n-1):
    a, b = map(int, input().split())
    g[a].append(b)
    g[b].append(a)
for i in range(1, n+1):
    if len(g[i]) == 1:
        root = i
        break
dep, stk = [0]*(n+1), [(root, 0)] #optim pt maxim daca pun radacina intr-o frunza
while len(stk):
    nod, t = stk[-1]
    dep[nod] = 1 + dep[t]
    stk.pop()
    for nn in g[nod]:
        if nn != t:
            stk.append((nn, nod))
frunze = []
for i in range(1, n+1):
    if len(g[i]) == 1:
        frunze.append(dep[i])
frunze.sort()
i = 0
while i < len(frunze) and (i == 0 or ~(frunze[i]-frunze[i-1])&1):
    i += 1
if i >= len(frunze):
    print(1, end = ' ') #umplu toate muchiile cu 1, pt ca toate drumurile dintre frunze au lg para
else:
    print(3, end = ' ') #1^2^3 == 0, poti sparge muchiile in 1^2^1^2^... si 1^2^3^...
m = n-1 - len(frunze) + 1 #numar de muchii care nu sunt "terminale"
#pe muchiile terminale pui xor-ul valorilor de la radacina pana la ea (2^??) ai sa anulezi ce ai facut
#astfel, xorpath(R, frunza) == 0 pt oricare frunza
#!!mai mult xorpath(frunza, frunza) == 0 chiar daca lca(frunze) != R
for i in range(1, n+1):
    if dep[i] > 2 and len(g[i]) > 1:
        add = 0
        for nn in g[i]:
            add |= (len(g[nn]) == 1)
        m += add #adaug muchiile terminale distincte
print(m)