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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second

int n;
vector<int> e[100010];

int par[100010];

int X;

int dfs(int a){
	vector<int> v;
	for(int b:e[a]){
		if(par[a] == b) continue;
		par[b]=a;
		v.pb(dfs(b)+1);
	}
	sort(all(v));
	int cc = v.size();
	if(a == 1){
		if(cc % 2){
			if(v[cc-1] > X) throw 1;
			for(int l=0,r=cc-2; l<r; ++l,--r) if(v[l]+v[r]>X) throw 1;
			return 0;
		} else {
			for(int l=0,r=cc-1; l<r; ++l,--r) if(v[l]+v[r]>X) throw 1;
			return 0;
		}
	} else if(e[a].size() % 2){
		bool just = true;
		for(int l=0, r=cc-1; l<r; ++l, --r){
			if(v[l] + v[r] > X){
				just = false;
				break;
			}
		}
		if(just){
			return 0;
		}
		if(v[cc-1] > X) throw 1;
		for(int l=1, r=cc-2; l<r; ++l, --r){
			if(v[l] + v[r] > X){
				just = false;
				break;
			}
		}
		return v[0];
	} else {
		for(int l=0, r=cc-2; l<r; ++l, --r) if(v[l] + v[r] > X) throw 1;
		int l=-1, r=cc-1;
		while(l+1 < r){
			int mid = (l+r)/2;
			bool fok = 1;
			for(int pl=0, pr=cc-1; pl<pr;){
				if(pl == mid){ ++pl; continue; }
				if(pr == mid){ --pr; continue; }
				if(v[pl] + v[pr] > X){
					fok = 0;
					break;
				}
				++pl; --pr;
			}
			if(fok) r=mid;
			else l=mid;
		}
		return v[r];
	}
}

bool check(int x){
	X = x;
	fill(par+1, par+n+1, 0);
	try {
		dfs(1);
		return 1;
	} catch(int t){ return 0; }
}

int main()
{
	read(n);
	for(int i=1; i<n; ++i){
		int a, b; read(a, b);
		e[a].pb(b); e[b].pb(a);
	}
	int oc = count_if(e+1, e+n+1, [](auto& v){ return v.size()%2; });
	printf("%d ", oc/2);
	
	int L = 0, R = n-1;
	while(L+1 < R){
		int mid = (L+R)/2;
		if(check(mid)) R = mid; else L = mid;
	}
	
	printf("%d\n", R);
	return 0;
}