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>
# -*- coding:utf-8 -*-

"""

created by shuangquan.huang at 2020/6/30

"""

import collections
import time
import os
import sys
import bisect
import heapq
from typing import List


# def dfs(G, node, parent, dist):
#     if not node:
#         return dist % 2 == 0
#
#     isleaf = True
#     for u in G[node]:
#         if u != parent:
#             isleaf = False
#             if not dfs(G, u, node, dist+1):
#                 return False
#
#     return dist % 2 == 0 if isleaf else True
    

if __name__ == '__main__':
    N = int(input())
    G = [[] for _ in range(N+1)]
    
    for i in range(N-1):
        u, v = map(int, input().split())
        G[u].append(v)
        G[v].append(u)
    
    leafs = [i for i in range(1, N+1) if len(G[i]) == 1]
    leafNeibs = set()
    for u in leafs:
        for v in G[u]:
            leafNeibs.add(v)
    
    maxv = (N - 1) - len(leafs) + len(leafNeibs)
    minv = 1
    
    q = [leafs[0]]
    dist = [N+1 for _ in range(N+1)]
    d = 0
    dist[leafs[0]] = 0
    while q:
        nq = []
        d += 1
        for u in q:
            for v in G[u]:
                if dist[v] == N+1:
                    dist[v] = d
                    nq.append(v)
        q = nq
    
    if any([dist[v] % 2 != 0 for v in leafs]):
        minv = 3
            
    print('{} {}'.format(minv, maxv))
