Problem: Takahashi has decided to make a Christmas Tree for the Christmas party in AtCoder, Inc.

A Christmas Tree is a tree with N vertices numbered 1 through N and N-1 edges, whose i-th edge (1\leq i\leq N-1) connects Vertex a_i and b_i.

He would like to make one as follows:

* Specify two non-negative integers A and B.
* Prepare A Christmas Paths whose lengths are at most B. Here, a Christmas Path of length X is a graph with X+1 vertices and X edges such that, if we properly number the vertices 1 through X+1, the i-th edge (1\leq i\leq X) will connect Vertex i and i+1.
* Repeat the following operation until he has one connected tree:
* Select two vertices x and y that belong to different connected components. Combine x and y into one vertex. More precisely, for each edge (p,y) incident to the vertex y, add the edge (p,x). Then, delete the vertex y and all the edges incident to y.
* Properly number the vertices in the tree.



Takahashi would like to find the lexicographically smallest pair (A,B) such that he can make a Christmas Tree, that is, find the smallest A, and find the smallest B under the condition that A is minimized.

Solve this problem for him.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq a_i,b_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

For the lexicographically smallest (A,B), print A and B with a space in between.

Examples

Input

7
1 2
2 3
2 4
4 5
4 6
6 7


Output

3 2


Input

8
1 2
2 3
3 4
4 5
5 6
5 7
5 8


Output

2 5


Input

10
1 2
2 3
3 4
2 5
6 5
6 7
7 8
5 9
10 5


Output

3 4

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
#define mod 1000000007
#define FOR(x,to) for(int x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define long long long
inline int rei(){int x;cin>>x;return x;}
inline long rel(){long x;cin>>x;return x;}
inline string res(){string x;cin>>x;return x;}
//------------------------------------------------------- 
vector<int> G[100000];
int ar[100000];
int buffer[100000];
int N;
bool check(int B){
	stack<pair<int,int>> sp;
	sp.push({0,-1});
	while(!sp.empty()){
		int v = sp.top().first;
		int f = sp.top().second;
		sp.pop();
		if(v >= N){
			v -= N;
			int p = 0;
			for(int i=0;i<G[v].size();i++){
				int t = G[v][i];
				if(t != f){
					buffer[p++] = ar[t]+1;
				}
			}
			if(v == 0){
				if(p % 2 == 1){
					buffer[p++] = 0;
				}
				sort(buffer,buffer+p);
				int l = 0;
				int r = p-1;
				while(l < r){
					if(buffer[l] + buffer[r] > B){
						return false;
					}
					l++;
					r--;
				}
			}
			else{
				if(p % 2 == 0){
					buffer[p++] = 0;
				}
				sort(buffer,buffer+p);
				int bf = 0;
				int bl = p;
				while(bf != bl){
					int bc = (bf+bl)/2;
					bool OK = true;
					if(bc != p){
						int l = 0;
						int r = p-1;
						while(l < r){
							if(l == bc){
								l++;
							}
							if(r == bc){
								r--;
							}
							if(buffer[l] + buffer[r] > B){
								OK = false;
								break;
							}
							l++;
							r--;
						}
					}
					if(OK){
						bl = bc;
					}
					else{
						bf = bc+1;
					}
				}
				if(bf == p){
					return false;
				}
				ar[v] = buffer[bf];
			}
		}
		else{
			sp.push({v+N,f});
			for(int i=0;i<G[v].size();i++){
				int t = G[v][i];
				if(t != f){
					sp.push({t,v});
				}
			}
		}
	}
	return true;
}
void Calc(){
	N = rei();
	for(int i=0;i<N-1;i++){
		int a = rei()-1;
		int b = rei()-1;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int A = 0;
	stack<pair<int,int>> sp;
	sp.push({0,-1});
	while(!sp.empty()){
		int v = sp.top().first;
		int f = sp.top().second;
		sp.pop();
		int b = 0;
		for(int i=0;i<G[v].size();i++){
			int t = G[v][i];
			if(t != f){
				b++;
				sp.push({t,v});
			}
		}
		A -= b / 2;
		if(b % 2 == 0 && f != -1){
			A++;
		}
	}
	int bf = 1;
	int bl = N-1;
	while(bf != bl){
		int bc = (bf+bl)/2;
		if(check(bc)){
			bl = bc;
		}
		else{
			bf = bc+1;
		}
	}
	cout << A << " " << bf << endl;
}
int main(int argc,char** argv){
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0); Calc(); return 0;
}