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>
using namespace std;
#define REP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i<=i##end;++i)
#define DREP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i>=i##end;--i)
template<typename T>bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
#ifdef __linux__
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif
template<typename T>T read(){
	T x=0,f=1;
	char c=getchar();
	while((c<'0')||(c>'9')){if(c=='-')f=-1;c=getchar();}
	while((c>='0')&&(c<='9'))x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
#define read() read<int>()
template<typename T>void write(T x,char c){
	static char t[25];
	static int tlen;
	t[tlen=1]=c;
	if(x<0)putchar('-'),x=-x;
	do t[++tlen]=(x%10)^48;
	while(x/=10);
	while(tlen)putchar(t[tlen--]);
}
#define pb push_back
typedef long long ll;
typedef double lf;
const int maxn=100005;
vector<int>E[maxn];
int n,deg[maxn],dp[maxn],lim,rt;
bool dfs(int u,int fa){
	bool res=1;
	REP(i,0,E[u].size()-1){
		int v=E[u][i];
		if(v!=fa)res&=dfs(v,u);
	}
	static int c[maxn],tot;
	tot=0;
	REP(i,0,E[u].size()-1){
		int v=E[u][i];
		if(v!=fa)c[++tot]=dp[v];
	}
	if(deg[u]&1)c[++tot]=0;
	sort(c+1,c+1+tot);
	int l=1,h=tot;
	while(l<=h){
		int mid=(l+h)>>1,curl=1,curr=tot,Max=0;
		REP(i,1,tot>>1){
			if(curl==mid)curl++;
			if(curr==mid)curr--;
			chkmax(Max,c[curl]+c[curr]);
			curl++,curr--;
		}
		if(Max<=lim)h=mid-1;
		else l=mid+1;
	}
	res&=(l<=tot);
	dp[u]=c[l]+1;
	return res;
}
bool check(int x){
	lim=x;
	return dfs(rt,0)&(dp[rt]-1<=lim);
}
int main(){
	int u,v;
	n=read();
	REP(i,1,n-1){
		u=read(),v=read();
		E[u].pb(v),deg[u]++;
		E[v].pb(u),deg[v]++;
	}
	int ans=0;
	REP(i,1,n)
		if(deg[i]&1)rt=i,ans++;
	write(ans>>1,' ');
	int l=1,h=n-1;
	while(l<=h){
		int mid=(l+h)>>1;
		if(check(mid))h=mid-1;
		else l=mid+1;
	}write(l,'\n');
	return 0;
}