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;

#define rep(i,l,r) for(int i=l;i<=r;++i)
#define pb push_back
const int N=1e5+5;
int n;
vector<int>lk[N],son[N];

int A;
namespace GET_A
{
void dfs(int x,int fr)
{
	for(auto y:lk[x])
	if(y!=fr)
	{
		son[x].pb(y);
		dfs(y,x);
	}
//	cerr<<x<<" "<<son[x].size()<<endl;
//	for(auto y:son[x])cerr<<y<<" ";puts("");
	if(x!=1)A+=son[x].size()/2;
	else A+=(son[x].size()+1)/2;
}
};
namespace GET_B
{
int len[N];
int mid;
bool dfs(int x)
{
	if(!son[x].size())
	{
		len[x]=1;
		return 0;
	}
	vector<int>a;
	for(auto y:son[x])
	{
		if(dfs(y))return 1;
		a.pb(len[y]);
	}
if(x==1)
{
	sort(a.begin(),a.end());
	int l=0,r=a.size()-1;
	if(r%2==0)
	{
		if(a[r]+1>mid)return 1;
		--r;
	}
	while(l<r)
	{
		if(a[l]+a[r]+1>mid)return 1;
		++l;--r;
	}
	return 0;
}
	if(a.size()%2==0)a.pb(0);
	sort(a.begin(),a.end());
	int l=0,r=a.size()-1;
	--r;
	while(l<r)
	{
		if(a[l]+a[r]+1>mid)return 1;
		++l;--r;
	}
	r=a.size()-1;l=0;
	while(r&&a[r]+a[l]+1<=mid)
	{
		--r;++l;
		if(l==r)++l;
	}
	len[x]=a[r]+1;
	return 0;
}
bool ok()
{
	if(dfs(1))return 0;
	return 1;
}
int erfen()
{
	int l=1,r=n;
	while(l+1!=r)
	{
		mid=(l+r)/2;
		if(ok())r=mid;
		else l=mid;
	}
//	mid=5;
//	ok();
	return r;
}
};

int main()
{
//	freopen("1.in","r",stdin);freopen("1.out","w",stdout);
	cin>>n;
	rep(i,1,n-1)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		lk[x].pb(y);lk[y].pb(x);
	}
	GET_A::dfs(1,0);
	int B=GET_B::erfen();
	cout<<A<<" "<<B-1<<endl;
}