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 <cstdio>
#include <algorithm>
#include <vector>

using std::sort;
using std::vector;

const int MAXN=1e5+5;
int rt,n,A,B;
int he[MAXN],deg[MAXN];
int dp[MAXN];

struct line{int to,nex;}ed[MAXN<<1];

inline void addE(int u,int v){
	static int cnt;
	ed[++cnt]=(line){v,he[u]};
	he[u]=cnt;
}

inline bool cmp_dp(int a,int b){return dp[a]<dp[b];}

inline bool check(vector<int> &vec,int son){
	for(int l=0,r=vec.size()-1;l<r;++l,--r){
		if(l==son) ++l;
		if(r==son) --r;
		if(dp[vec[l]]+dp[vec[r]]>B)
			return false;
	}
	return true;
}

inline int bin_chop_son(vector<int> &vec){
	int l=0,r=vec.size()-1,mid;
	while(l<=r){
		mid=l+r>>1;
		if(check(vec,mid)) r=mid-1;
		else l=mid+1;
	}
	return l;
}

bool treeDP(int u,int fa){
	bool ret=true;
	vector<int> vec;
	for(int i=he[u],v;i;i=ed[i].nex){
		v=ed[i].to;
		if(v!=fa){
			ret&=treeDP(v,u);
			vec.push_back(v);
		}
	}
	if(!ret) return false;
	if(vec.size()==0){
		dp[u]=1;
		return true;
	}
	//度数为奇数，则儿子为偶数个
	if(deg[u]&1) vec.push_back(0);
	sort(vec.begin(),vec.end(),cmp_dp);
	int v=bin_chop_son(vec);
	if(v==vec.size()) return false;
	dp[u]=dp[vec[v]]+1;
	return dp[u]<=B+1;
}

inline void bin_chop_B(){
	int l=0,r=n;
	while(l<=r){
		B=l+r>>1;
		if(treeDP(rt,0)) r=B-1;
		else l=B+1;
	}
	B=l;
}

inline void cal_A(){
	int cnt=0;
	for(int i=1;i<=n;++i)
		cnt+=deg[i]&1;
	A=cnt>>1;
}

int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;++i){
		scanf("%d%d",&u,&v);
		++deg[u],++deg[v];
		addE(u,v),addE(v,u);
	}
	cal_A();
	for(int i=1;i<=n;++i){
		if(deg[i]==1){
			rt=i;
			break;
		}
	}
	bin_chop_B();
	printf("%d %d",A,B);
	return 0;
}