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 <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;

int n;
vector<int> G[100005];
bool used[100005];
P res[100005];
int cnt[100005];

P solve(int len){
	memset(used,false,sizeof(used));
	memset(cnt,0,sizeof(cnt));
	queue<int> que;
	for(int i=0;i<n;i++){
		if(G[i].size()==1){
			que.push(i);
		}
	}
	int prev=-1;
	while(que.size()){
		int v=que.front();
		prev=v;
		que.pop();
		used[v]=true;
		int sum=0;
		multiset<int> vi;
		for(int i=0;i<G[v].size();i++){
			int nv=G[v][i];
			if(used[nv]){
				vi.insert(res[nv].second);
				sum+=res[nv].first;
			}else{
				cnt[nv]++;
				if(cnt[nv]+1==G[nv].size())que.push(nv);
			}
		}
		if(vi.size()==0){
			res[v]=P(1,1);
			continue;
		}
		bool flag=false;
		int len_min=n+5;
		while(vi.size()){
			int v=*vi.rbegin();
			multiset<int>::iterator it2=vi.end();
			it2--;
			vi.erase(it2);
			multiset<int>::iterator it=vi.upper_bound(len-v);
			if(it!=vi.begin()){
				it--;
				vi.erase(it);
				sum--;
			}else{
				flag=true;
				len_min=min(len_min,v);
			}
		}
		if(!flag){
			sum++;
			len_min=0;
		}else if(len_min==len){
			sum++;
			len_min=0;
		}
		res[v]=P(sum,len_min+1);
		//printf("%d %d %d\n",v,res[v].first,res[v].second);
	}
	if(res[prev].second==1){
		res[prev].first--;
		res[prev].second=len;
	}
	return res[prev];
}


bool C(int num,int len){
	P rec=solve(len);
	if(rec.first>num)return false;
	return true;
}

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int l1=0,r1=n;
	while(l1+1<r1){
		int mid=(l1+r1)/2;
		if(C(mid,n))r1=mid;
		else l1=mid;
	}
	int l2=0,r2=n;
	while(l2+1<r2){
		int mid=(l2+r2)/2;
		if(C(r1,mid))r2=mid;
		else l2=mid;
	}
	printf("%d %d\n",r1,r2);
	return 0;
}