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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>

#define pb push_back
#define mp make_pair

using namespace std;

template<typename T>inline void upmin(T &x,T y) { y<x?x=y:0; }
template<typename T>inline void upmax(T &x,T y) { x<y?x=y:0; }

typedef unsigned int u32;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod pi=acos(-1);
const int oo=1<<30;
const LL OO=1e18;

const int N=1e6+100;

int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}

VI e[N];
int l,r,mid,f[N],n;
inline void dfs(int k) {
	for (int t:e[k]) {
		e[t].erase(find(e[t].begin(),e[t].end(),k));
		dfs(t);
	}
}
inline bool solve(int k) {
	int m=e[k].size(),i,L,R,MID;VI w(m);
	for (i=0;i<m;i++) {
		if (!solve(e[k][i])) return false;
		w[i]=f[e[k][i]]+1;
	}
	if (~m&1) w.pb(0),m++;
	sort(w.begin(),w.end());
	L=0,R=m;
	while (L!=R) {
		MID=(L+R)>>1;
		f[k]=w[MID];
		w.erase(w.begin()+MID);
		for (i=0;i<m/2;i++)
			if (w[i]+w[m-2-i]>mid) break;
		w.insert(w.begin()+MID,f[k]);
		if (i==m/2) R=MID;
		else L=MID+1;
	}
	if (L==m) return false;
	f[k]=w[L];
	return true;
}
int main()
{
	int n=gi(),i,a,b,ans=0,rt;
	for (i=1;i<n;i++) {
		a=gi(),b=gi();
		e[a].pb(b),e[b].pb(a);
	}
	for (i=1;i<=n;i++) ans+=e[i].size()&1,e[i].size()==1?rt=i:0;
	dfs(rt);
	l=1,r=n;
	while (l!=r) {
		mid=(l+r)>>1;
		if (solve(rt)&&f[e[rt][0]]<mid) r=mid;
		else l=mid+1;
	}
	cout<<ans/2<<" "<<l<<endl;
	return 0;
}