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 math
import sys
from collections import defaultdict,Counter,deque,OrderedDict
import bisect
#sys.setrecursionlimit(1000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
#def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#INF = 10 ** 18
#MOD = 1000000000 + 7
from itertools import accumulate,groupby
import sys
import heapq

first = 1 #if first node is 1
G = defaultdict(list)
def addEdge(a,b):
    G[a].append(b)
    G[b].append(a)
	
def BFSUtil(queue,vis,level):
    while queue:
        s = queue.popleft()
        for child in G.get(s,[]):
            if vis[child] == False:
                queue.append(child)
                vis[child] = True
                level[child] = level[s] + 1 
            
def sssp(node,N):
    vis = [False for i in range(N+1)]
    level = [0 for i in range(N+1)]
    queue = deque()
    queue.append(node)
    vis[node] = True
    level[node] = 0 
    BFSUtil(queue,vis,level)
    return level
	
N = int(input())
maxi = N-1
for _ in range(N-1):
    u,v = ilele()
    addEdge(u,v)
if N==3:
    maxi,mini = 1,1
else:
    l = {};source =None
    for i,j in G.items():
        if len(j) == 1:
            l[i] = 1
            source = i
    X  = sssp(source,N)
    #print(X,l,G)
    f= 0
    for i in l.keys():
        if X[i] % 2!= 0:
            f=1
            break
    #print(f)
    nodemark ={}
    for i,j in G.items():
        if len(j)==1:
            if nodemark.get(j[0],-1) == 1:
                maxi -= 1
            else:
                nodemark[j[0]] = 1
    if f==0:
        mini = 1
    else:
        mini = 3
print(mini,maxi)    
        
