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
 
typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef unsigned long long ull;
typedef long double ld; 
typedef tree<ii, null_type, less<ii>, rb_tree_tag, tree_order_statistics_node_update> pbds;

int n; 
const int INF = int(1e9);
const int N = 123456;
ii dp[N];
vi adj[N];

int threshold;

void dfs(int u, int p=-1)
{
	int sum = 0;
	deque<int> V;
	for(int v:adj[u])
	{
		if(v==p) continue;
		dfs(v,u);
		sum+=dp[v].fi;
		V.pb(dp[v].se);
	}
	sort(V.begin(),V.end());
	while(!V.empty()&&V.back()>=INF)
	{
		V.pop_back(); V.push_front(0); sum++;
	}
	int mnup=INF;
	while(!V.empty())
	{
		if(V.size()>=2&&V[0]+V.back()+2<=threshold)
		{
			sum--; V.pop_front(); V.pop_back(); continue;
		}
		if(V.back()+1<=threshold)
		{
			mnup=min(mnup,V.back()+1); V.pop_back(); continue;
		}
		mnup=min(mnup,1); V.pop_back(); sum++;
	}
	dp[u] = mp(sum,mnup);
	//cerr<<threshold<<" | "<<u<<' '<<sum<<' '<<mnup<<'\n';
}

int calc(int C)
{
	threshold=C;
	dfs(0,-1); 
	return dp[0].fi;
}

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	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);
	}
	int bestres = calc(n+1);
	int lo=1; int hi=n;
	int ans=-1;
	while(lo<=hi)
	{
		int mid=(lo+hi)>>1;
		int v=calc(mid);
		if(v<=bestres)
		{
			ans=mid; hi=mid-1;
		}
		else
		{
			lo=mid+1;
		}
	}
	cout<<bestres<<' '<<ans<<'\n';
}