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<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#define N 200010
using namespace std;
int n,con,h[N<<1],in[N],Rt,f[N];
struct edge
{
	int to,ne;
}e[N<<1];
int read()
{
	int x=0,f=1; 
	char ch=getchar();
	while(ch<'0'||ch>'9')
	{
		if(ch=='-')
			f=-1; 
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')
	{
		x=x*10+ch-'0'; 
		ch=getchar();
	}
	return x*f;
}
void add(int u,int v)
{
	e[++con].to=v;
	e[con].ne=h[u];
	h[u]=con;
}
void dfs(int rt,int pr,int x)
{
	vector<int>v; 
	v.clear();
	for(int i=h[rt];i;i=e[i].ne)
	if(e[i].to!=pr) 
	{
		dfs(e[i].to,rt,x);
		v.push_back(f[e[i].to]);
	}
	if(!(v.size()&1)) 
		v.push_back(0);
	sort(v.begin(),v.end());
	int l=0,r=v.size()-1,rs=n+1;
	while(l<=r)
	{
		int mid=(l+r)>>1,fg=1;
		for(int i=0,j=v.size()-1;i<j;i++,j--)
		{
			if(i==mid) 
				i++; 
			if(j==mid) 
				j--;
			fg&=(v[i]+v[j])<=x;
		}
		if(fg) 
			
		r=mid-1,rs=v[mid];
		else l=mid+1;
	}
	f[rt]=rs+1;
}
bool check(int x)
{
	dfs(Rt,0,x); 
	return f[Rt]<=x+1;
}
int main()
{
	n=read(); 
	for(int i=1;i<n;i++)
	{
		int u=read(),v=read();
		add(u,v);
		in[u]++;
		add(v,u);
		in[v]++;
	}
	int cnt=0; 
	for(int i=1;i<=n;i++) 
		cnt+=in[i]&1,Rt=in[i]==1?i:Rt;
	int l=0,r=n,ans=0;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(check(mid)) 
			r=mid-1,ans=mid;
		else 
			l=mid+1;
	}
	cout<<(cnt+1)/2<<" "<<ans<<endl;
	return 0;
}