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 os
import sys
input = sys.stdin.buffer.readline
#sys.setrecursionlimit(int(3e5)) 
from collections import deque
from queue import PriorityQueue
import math
#  list(map(int, input().split()))
#####################################################################################

class CF(object):    
    def __init__(self):
        self.n = int(input())
        self.g = [[] for _ in range(self.n)]
        for _ in range(self.n - 1):
            x,y = list(map(int, input().split()))
            x-=1
            y-=1
            self.g[x].append(y)
            self.g[y].append(x)
        pass
        self.vis = [0] * self.n
        self.f = [-1] * self.n
        self.h = [0] * self.n
        self.leaf = []
        for i in range(self.n):
            if(len(self.g[i]) == 1):
                self.leaf.append(i)
        pass
        self.root = self.leaf[0]
    
    def bfs(self):
        que = deque([self.root])                                
        self.vis[self.root] = True
        while(que):
            now = que.popleft()
            for to in self.g[now]:                
                if(not self.vis[to]):
                    self.h[to] = self.h[now] + 1
                    self.vis[to] = True
                    self.f[to] = now
                    que.append(to)
            pass
        pass

    def check(self):
        for i in range(1, len(self.leaf)):
            if(self.h[self.leaf[i]] % 2 != 0):
                return False
        pass
        return True

    def main(self):                    
        self.bfs()
        ans_min = None
        if(self.check()):
            ans_min = 1
        else:
            ans_min = 3

        ans_max = self.n - 1
        cnt = 0
        self.vis = [0] * self.n
        for i in range(1, len(self.leaf)):
            if(self.h[self.leaf[i]] == 2):
                cnt += 1
            else:
                if(self.vis[self.f[self.leaf[i]]]):
                    cnt+=1
            pass
            self.vis[self.f[self.leaf[i]]] = True
        pass
        ans_max = ans_max - cnt
        print(str(ans_min) + ' '+ str(ans_max))


if __name__ == "__main__":
    cf = CF()
    cf.main()
    pass
    


