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>
n = int(input())
graph = [[] for i in range(n+1)]
deg = [0]*(n+1)

for i in range(n-1):
    x,y = map(int,input().split())
    graph[x].append(y)
    graph[y].append(x)
    deg[x]+=1
    deg[y]+=1
root =-1
leavs = []
for i in range(n+1):
    if(deg[i]==1):
        root =i
        break

for i in range(n+1):
    if(deg[i]==1):
        leavs.append(i)
# min_ans =1

stack = []
depth = [0]*(n+1)
stack.append(root)
visi = set()

visi.add(root)
cnt =0
gg = 0
nod = graph[root][0]
while(len(stack)!=0):
    x = stack.pop()
    fn = 0
    for nodes in graph[x]:
        if(nodes not in visi):
            stack.append(nodes)
            visi.add(nodes)
            depth[nodes] = depth[x]+1
            if(deg[nodes]==1):
                fn=1
    if(fn):
        cnt+=1
    if(x == nod and fn==1):
        gg=1

fl= 0
for i in leavs:
    if(depth[i]%2):
        fl=1
if(fl):
    print(3,end=" ")
else:
    print(1,end =" ")

if(gg==0):
    cnt+=1

print(n+cnt - 1 - (len(leavs)))
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