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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const double EPS=1e-6;
const double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
vector<int>g[110000];
bool dame=false;
int M;
int v[110000];
int dfs(int a,int b){
	vector<int>tmp;
	int sz=0;
	for(int i=0;i<g[a].size();i++){
		if(b==g[a][i])continue;
		sz++;
		int t=dfs(g[a][i],a);
		if(t>M)dame=true;
		if(dame)return 0;
		tmp.push_back(t);
	}
	std::sort(tmp.begin(),tmp.end());
	if(b==-1){
		for(int i=0;i<tmp.size()/2;i++){
			if(tmp[i]+tmp[tmp.size()-1-i]>M)dame=true;
		}
		return 0;
	}else if(sz%2==0){
		bool ok=true;
		for(int i=0;i<tmp.size()/2;i++){
			if(tmp[i]+tmp[tmp.size()-1-i]>M)ok=false;
		}
		if(ok)return 1;
		int cnt=0;
		for(int i=0;i+1<tmp.size()/2;i++){
			v[i]=tmp[i+1]+tmp[tmp.size()-2-i];
			if(v[i]>M)cnt++;
		}
		int at=-1;
		if(cnt==0)at=0;
		for(int i=1;at==-1&&i+1<tmp.size();i++){
			int x=i-1;
			if(1+x>=tmp.size()/2)x=tmp.size()/2-2-(x-tmp.size()/2+1);
			if(v[x]>M&&v[x]+tmp[i-1]-tmp[i]<=M)cnt--;
			v[x]+=tmp[i-1]-tmp[i];
			if(cnt==0){
				at=i;break;
			}
		}
		if(at!=-1)ok=true;
		if(ok)return tmp[at]+1;
		else dame=true;
		return 0;
	}else{
		int cnt=0;
		for(int i=0;i<tmp.size()/2;i++){
			v[i]=tmp[i+1]+tmp[tmp.size()-1-i];
			if(v[i]>M)cnt++;
		}
		int at=-1;
		if(cnt==0)at=0;
		for(int i=1;at==-1&&i<tmp.size();i++){
			int x=i-1;
			if(x>=tmp.size()/2)x=tmp.size()/2-1-(x-tmp.size()/2);
			if(v[x]>M&&v[x]+tmp[i-1]-tmp[i]<=M)cnt--;
			v[x]+=tmp[i-1]-tmp[i];
			if(cnt==0){
				at=i;break;
			}
		}
		if(at==-1){dame=true;return 0;}
		return tmp[at]+1;
	}
}
int main(){
	int a;scanf("%d",&a);
	for(int i=0;i<a-1;i++){
		int p,q;scanf("%d%d",&p,&q);p--;q--;
		g[p].push_back(q);g[q].push_back(p);
	}
	int req=1;
	for(int i=0;i<a;i++){
		if(g[i].size()>2)req+=(g[i].size()-1)/2;
	}
	int left=0;
	int right=a;
	while(left+1<right){
		M=(left+right)/2;
		dame=false;
		dfs(0,-1);
		if(dame==false){
			right=M;
		}else left=M;
	}
	printf("%d %d\n",req,right);
}