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>

using namespace std;

#define For(i,a,b) for(int i = a; i <= b; ++ i)
#define pb push_back
#define fi first
#define se second

const int N = 1e5 + 5;
typedef pair < int, int > ii;
int t, n, h[N];
bool kt;
vector < int > adj[N];

bool check1(vector < ii > data, int pos, int lim){
	if(pos >= 0)data.erase(data.begin() + pos);
	int Max = 0, len = data.size();
	For(i, 0, len / 2 - 1) Max = max(Max, data[i].fi + data[len - i - 1].fi);
	if (Max > lim) return false;
	return true;
}
int cal(vector < ii > data, int lim){
	int len = data.size();
	int l = 0, r = len - 1, res = len;
	while(l <= r){
		int mid = (l + r) / 2;
		if(check1(data, mid, lim)) res = mid, r = mid - 1;
		else l = mid + 1;
	}
	if(res < len) return data[res].se;
	return -1;
}
void dfs(int u, int pa, int lim){
	if(! kt) return; 
	int child = 0; vector < ii > data;
	For(i, 0, (int)adj[u].size() - 1){
		int v = adj[u][i]; if(v == pa) continue;
		if(! kt) return;
		child++, dfs(v, u, lim);
		if(h[v] + 1 > lim) { kt = false; return; }
		data.pb(ii(h[v] + 1, v));
	}
	if(!child) return;
	sort(data.begin(), data.end());
	int len = data.size(), Max = 0; 
	if(len % 2 == 0){
		kt = check1(data, -1, lim);
		if(!kt) return;
	}
	else{
		int res = cal(data, lim);
		if(res == -1) { kt = false; return; }
		h[u] = h[res] + 1;
		if(u == 1 && h[u] > lim) { kt = false; return; }
	}
}
bool check(int mid){
	memset(h, 0, sizeof h);
	kt = true; dfs(1, 1, mid);
	return kt;
}
signed main(){
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n;
	For(i, 1, n - 1){
		int u, v;
		cin >> u >> v;
		adj[u].pb(v);  adj[v].pb(u);
	}
	int dem = 0;
	For(i, 1, n) if(adj[i].size() % 2) dem ++; dem /= 2;
	int l = 0, r = n, ans;
	while(l <= r){
		int mid = (l + r)/2;
		if(check(mid)) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	cout << dem  << ' ' << ans;
}
/*
6
1 2
1 3
1 4
1 5
1 6
*/