Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

7
1 2
2 3
2 4
4 5
4 6
6 7

Sample Output 1

3 2

We can make a Christmas Tree as shown in the figure below:

Sample Input 2

8
1 2
2 3
3 4
4 5
5 6
5 7
5 8

Sample Output 2

2 5

Sample Input 3

10
1 2
2 3
3 4
2 5
6 5
6 7
7 8
5 9
10 5

Sample Output 3

3 4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iomanip>
#include<set>
#define sqr(x) (x)*(x)
using namespace std;
int n,m,i,j,ans,ans2=n+n,l,r,mid,x,y,dis[100005];
vector<int> bi[100005];
bool dfs(int x,int fa)
{
	int i;
	vector<int> len;
	for (i=0;i<bi[x].size();i++)
	{
		if (bi[x][i]==fa) continue;
		if (!dfs(bi[x][i],x)) return 0;
		len.push_back(dis[bi[x][i]]+1);
	}
	if (len.size()%2==1) len.push_back(0);
	sort(len.begin(),len.end());
	set<int> s;
	for (i=0;i<len.size();i++) s.insert(i);
	int j=len.size()-1;i=0;
	int t=len.size()/2;
	while (i<j&&t)
	{
		while (t&&i<=j&&len[i]+len[j]>mid)
		{
			if (len[j]<=mid)
			{
				s.erase(j);
				t--;
				j--;
			}
			else return 0;
		}
		if (i>=j||!t) break;
		s.erase(i);
		s.erase(j);
		i++;
		j--;
		t--;
	}
	if (s.size()>1) return 0;
	if (s.size()==1)
	{
		if (x==1) return 0;
		dis[x]=len[*s.begin()];
	}
	return 1;
}
bool check(int x)
{
	int i;
	memset(dis,0,sizeof(dis));
	if (!dfs(1,0)) return 0;
	if (dis[1]>x) return 0;
	return 1;
}
int main()
{
	cin>>n;
	for (i=1;i<n;i++)
	{
		cin>>x>>y;
		bi[x].push_back(y);
		bi[y].push_back(x);
	}
	ans=n-1;
	for (i=1;i<=n;i++)
	{
		ans-=bi[i].size()/2;
	}
	l=0;r=n+n;
	while (l<r)
	{
		mid=(l+r+1)/2;
		if (check(mid))
		{
			r=mid-1;
			ans2=mid;
		}
		else l=mid;
	}
	cout<<ans<<' '<<ans2<<endl;
	return 0;
}