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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int n;
vvi g;
vi dp;

bool ff(vi& a,int I,int L){
	int N=a.size();
	bool B=1;
	for(int i=0;i<(N-1)/2;i++) if(a[(i<I?i:i+1)]+a[(N-i-1>I?N-i-1:N-i-2)]>L) B=0;
	return B;
}

bool f(int v,int p,int L){
	bool B=1;
	vi a;
	for(auto u:g[v]) if(u!=p){
		B&=f(u,v,L)&&dp[u]!=L;
		a.push_back(dp[u]+1);
	}
	if(!B) return 0;
	sort(a.begin(),a.end());
	int N=a.size();
	if(N%2==0){
		for(int i=0;i<N/2;i++) if(a[i]+a[N-i-1]>L) B=0;
		if(B) return 1;
		a.pop_back();
		N--;
	}
	int l=-1,r=N;
	while(r-l>1){
		int m=(l+r)/2;
		if(ff(a,m,L)) r=m;
		else l=m;
	}
	if(r==N) return 0;
	dp[v]=a[r];
	return 1;
}

int main(){
	cin>>n;
	g=vvi(n);
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	int A=0,v,l=0,r=n-1;
	for(int i=0;i<n;i++){
		int S=g[i].size();
		A+=S%2;
		if(S==1) v=i;
	}
	A/=2;
	while(r-l>1){
		dp=vi(n);
		int m=(l+r)/2;
		if(f(v,-1,m)) r=m;
		else l=m;
	}
	dp=vi(n);
	cout<<A<<' '<<r<<endl;
}