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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key

typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef long double ld; 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;


ii dp[111111];
vi adj[111111];
int MID;
int tot;
void dfs(int u, int p)
{
	int cnt=0;
	deque<int> vec;
	for(int i=0;i<adj[u].size();i++)
	{
		int v=adj[u][i]; if(v==p) continue;
		dfs(v,u);
		vec.pb(dp[v].se);
		cnt+=dp[v].fi;
	}
	sort(vec.begin(),vec.end());
	/*
	int maxpath = 0;
	while(!vec.empty())
	{
		if(vec.size()==1)
		{
			if(vec[0]+1<=mid)
			{
				maxpath = max(maxpath, vec[0]+1);
			}
			else
			{
				cnt++;
				maxpath = max(maxpath, 1);
			}
			vec.pop_back();
		}
		else
		{
			int las = vec.back();
			if(vec.front()+2+las<=mid)
			{
				
			}
			else
			{
				cnt++;
				vec.pop_back();
				vec.push_front(0);
			}
		}
	}
	*/
	int lo = 0; int hi = int(vec.size()); int ans = 0; //[ans, int(vec.size()) - 1] will be killed
	while(lo<=hi)
	{
		int mid=(lo+hi)>>1;
		int maxi = -int(1e9);
		for(int i=0;i<mid;i++)
		{
			if(i<mid-1-i) maxi=max(maxi,vec[i]+vec[mid-1-i]+2);
			maxi=max(maxi,vec[i]+1);
		}
		if(maxi<=MID)
		{
			ans=mid;
			lo=mid+1;
		}
		else hi=mid-1;
	}
	int maxpath = 0;
	int del=0;
	for(int i=ans;i<vec.size();i++) 
	{
		if(vec[i]+1<=MID)
		{
			del++;
		}
		vec[i]=0;
	}
	cnt+=int(vec.size()) - ans;
	for(int j=0;j<del;j++) vec.pop_back();
	sort(vec.begin(),vec.end());
	//cerr<<vec.size()<<'\n';
	cnt+=int(vec.size())/2;
	if(vec.size()%2==1)
	{
		lo = 0; hi = int(vec.size())-1; ans=0;
		while(lo<=hi)
		{
			int mid=(lo+hi)>>1;
			int l = 0; int r = int(vec.size())-1;
			int res = -int(1e9);
			while(l==mid) l++;
			while(r==mid) r--;
			while(l<r)
			{
				res=max(res,vec[l]+vec[r]+2);
				l++; r--;
				while(l==mid) l++;
				while(r==mid) r--;
			}
			if(res<=MID)
			{
				ans=mid; hi=mid-1;
			}
			else
			{
				lo=mid+1;
			}
		}
		maxpath = vec[ans]+1;
		if(maxpath>MID)
		{
			maxpath=0;
			cnt++;
		}
	}
	//cerr<<MID<<' '<<u+1<<' '<<cnt<<' '<<maxpath<<'\n';
	dp[u] = mp(cnt,maxpath);
}

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	int n; cin>>n;
	for(int i=0;i<n-1;i++)
	{
		int u,v; cin>>u>>v; u--; v--;
		adj[u].pb(v); adj[v].pb(u);
	}
	for(int i=0;i<n;i++)
	{
		tot+=(adj[i].size()&1);
	}
	MID=n;
	dfs(0,-1);
	int res=dp[0].fi;
	if(dp[0].se>0) res++;
	assert(res==tot/2);
	int lo=1; int hi=n; int ans=-1;
	while(lo<=hi)
	{
		MID=(lo+hi)>>1;
		dfs(0,-1);
		int res2=dp[0].fi;
		if(dp[0].se>0) res2++;
		if(res2==res)
		{
			ans=MID; hi=MID-1;
		}
		else lo=MID+1;
	}
	cout<<res<<' '<<ans<<'\n';
}