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>
#define N 200010
using namespace std;
 
inline 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;
}
struct edge{int v,next;}vs[N<<1];
int n,ee,st[N<<1],du[N],Rt,F[N];
inline void addedge(int u,int v)
{
	vs[++ee].v=v;vs[ee].next=st[u];st[u]=ee;
}
inline void dfs(int rt,int pr,int x)
{
	vector <int> v; v.clear();
	for(int i=st[rt];i;i=vs[i].next)
	{
		if(vs[i].v==pr) continue;
		dfs(vs[i].v,rt,x);
		v.push_back(F[vs[i].v]);
	}
	if(!(v.size()&1)) v.push_back(0);
	sort(v.begin(),v.end());
	// printf("son of node %d len of %d\n",rt,x);
	// for(int i=0;i<v.size();i++) printf("%d ",v[i]); puts("");
	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;
}
inline bool check(int x)
{
	dfs(Rt,0,x); return F[Rt]<=x+1;
}
int main()
{
	// freopen("read.in","r",stdin);
	n=read(); for(int i=1;i<n;i++)
	{
		int u=read(), v=read();
		addedge(u,v); du[u]++;
		addedge(v,u); du[v]++;
	}
	int cnt=0; for(int i=1;i<=n;i++) 
		cnt+=du[i]&1,Rt=du[i]==1? i:Rt;
	//Rt=1;
	int l=1,r=n,B=0;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(check(mid)) r=mid-1,B=mid;
		else l=mid+1;
	}
	cout << cnt/2 << " " << B << endl;
	return 0;
}