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
#include<bits/stdc++.h>
using namespace std;

vector<int> edge[100001];
vector<int> v[100001];
int val,dp[100001];
bool bad;
int dfss(int x,int p){
	int res=0;
	for(int i:edge[x]){
		if(i==p)	continue;
		res+=dfss(i,x);
	}
	return (edge[x].size()-1)/2+res;
}

int dfs(int x,int p){
	v[x].clear();
	for(int i:edge[x]){
		if(i==p)	continue;
		v[x].push_back(dfs(i,x)+1);
	}
	sort(v[x].begin(),v[x].end());
	dp[x]=0;
	if(v[x].size()&1){
		int L=0,R=v[x].size();
		while(L<R){
			int M=(L+R)/2,st=0,en=v[x].size()-1;
			dp[x]=0;
			while(st<en){
				if(st==M){
					st++;
					continue;
				}
				if(en==M){
					en--;
					continue;
				}
				dp[x]=max(dp[x],v[x][st++]+v[x][en--]);
			}
			if(dp[x]>val)
				L=M+1;
			else
				R=M;
		}
		if(R==v[x].size()){
			bad=true;
			return 0;
		}
		else{
			if(v[x][L]>val)
				bad=true;
			return v[x][L];
		}
	}
	else{
		for(int i=0;i<v[x].size();i++)
			dp[x]=max(dp[x],v[x][i]+v[x][v[x].size()-i-1]);
		if(dp[x]>val){
			if(x==1){
				bad=true;
				return 0;
			}
			if(v[x].back()>val){
				bad=true;
				return 0;
			}
			int L=0,R=v[x].size()-1;
			while(L<R){
				int M=(L+R)/2,st=0,en=v[x].size()-2;
				dp[x]=0;
				while(st<en){
					if(st==M){
						st++;
						continue;
					}
					if(en==M){
						en--;
						continue;
					}
					dp[x]=max(dp[x],v[x][st++]+v[x][en--]);
				}
				if(dp[x]>val)
					L=M+1;
				else
					R=M;
			}
			if(R==v[x].size()-1){
				bad=true;
				return 0;
			}
			else{
				return v[x][L];
			}
		}
		else
			return 0;
	}
}

bool check(int x){
	bad=false;
	val=x;
	dfs(1,0);
	if(bad)
		return false;
	else
		return true;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0,u,v;i<n-1;i++){
		cin>>u>>v;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	int l=1,r=100000;
	while(l<r){
		int mid=(l+r)/2;
		if(check(mid))
			r=mid;
		else
			l=mid+1;
	}
	cout<<dfss(1,0)+1<<' '<<l<<endl;
}
