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())
G={}
for i in range(n-1):
    a,b=map(int,input().split())
    try:
        G[a].append(b)
    except:
        G[a]=[b]
    try:
        G[b].append(a)
    except:
        G[b]=[a]

leaves=[]
for i in G.keys():
    if (len(G[i])==1):
        leaves.append(i)
    else:
        sv=i
val=[-1 for i in range(n+1)]

Q=[sv]
val[sv]=0
tp=0
while(len(Q)!=0):
    a=Q.pop()
    iv=val[a]
    tc=-1
    for i in G[a]:
        if (val[i]==-1):
            val[i]=iv+1
            if (len(G[i])==1):
                tc=tc+1
            else:
                Q.append(i)
        
    tp=tp+max(tc,0) 
                
v=val[leaves[0]]
flag=0
for i in range(1,len(leaves)):
    if (abs(val[leaves[i]]-v)%2!=0):
        flag=1
        break
if (flag==0):
    print(1,end=" ")
else:
    print(3,end=" ")
    
print(n-tp-1)
            
