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<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
int n,cnt,ans,tot,q[100010],inf=1e9,l1,r1,mi,v[100010];
int x1[100010],r[200010],vt[200010],in[100010],mx,root;
void adde(int a,int b)
{
	r[++cnt]=x1[a];x1[a]=cnt;vt[cnt]=b;in[b]++;
	r[++cnt]=x1[b];x1[b]=cnt;vt[cnt]=a;in[a]++;
} 
void dfs(int i,int fa)
{
	int s,a,b,j,k;
	int l2,r2,m,st,ed;
	a=x1[i];v[i]=0;
	while(a)
	{
		b=vt[a];
		if(b!=fa)dfs(b,i);
		a=r[a];
	}
	a=x1[i];cnt=0;
	while(a)
	{
		b=vt[a];
		if(b!=fa)q[++cnt]=v[b]+1;
		a=r[a];
	}
	if(!(cnt&1))q[++cnt]=0;
	sort(q+1,q+1+cnt);
	l2=1;r2=cnt+1;
	while(l2<r2)
	{
		m=(l2+r2)/2;j=1;
		st=1;ed=cnt;
		while(st<ed)
		{
			if(st==m)st++;
			if(ed==m)ed--;
			if(q[st]+q[ed]>mi){j=0;break;}
			st++;ed--;
		}
		if(j)r2=m;
		else l2=m+1;
	} 
	if(l2==cnt+1)v[i]=inf;
	else v[i]=q[l2];
}
bool jd(int mi)
{
	dfs(root,0);
	return (v[root]<=mi);
}	
int main()
{
	int i,s,a,b,j,k;
	scanf("%d",&n);
	for(i=1;i<n;i++)scanf("%d%d",&a,&b),adde(a,b);
	ans=1;
	for(i=1;i<=n;i++)
	{
		ans+=(in[i]-1)/2;
		if(in[i]==1)root=i;
	}
	l1=1;r1=n;
	while(l1<r1)
	{
		mi=(l1+r1)/2;
		if(jd(mi))r1=mi;
		else l1=mi+1;
	}
	printf("%d %d",ans,l1);
	return 0;
}