Problem: 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.

Examples

Input

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


Output

3 2


Input

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


Output

2 5


Input

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


Output

3 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
					
using namespace std;
			
typedef long long ll;
#define mp make_pair
#define pub push_back
#define x first
#define y second
#define all(a) a.begin(), a.end()
#define db long double
 
int n;
vector<int> g[100007];
 
int calcCnt(int v, int pred){
	int ans = 0;
	int cnt = 0;
	for (auto to : g[v]) if (to != pred){
		ans += calcCnt(to, v);
		cnt++;
	}
	if (v != 0) cnt--;
	ans += (cnt + 1) / 2;
	return ans;
}
 
bool f;
int ww;
 
int dfs(int v, int pred){
	if (g[v].size() == 1 && g[v][0] == pred) return 0;
	vector<int> t;
	for (auto to : g[v]) if (to != pred){
		t.pub(dfs(to, v) + 1);
	}
	sort(all(t));
	if (t.back() > ww){
		f = 0;
		return 0;
	}
	if (v == 0){
		if (t.size() % 2 == 1) t.pop_back();
		for (int i = 0; i < t.size() / 2; i++){
			if (t[i] + t[(int)t.size() - i - 1] > ww){
				f = 0;
			}
		}
		return 0;
	} else {
		if (t.size() % 2 == 0){
			bool ok = 1;
			for (int i = 0; i < t.size() / 2; i++) if (t[i] + t[(int)t.size() - 1 - i] > ww) ok = 0;
			if (ok) return 0;
			t.pop_back();
		}
		int vl = -1, vr = t.size();
		while(vl + 1 < vr){
			int vm = (vl + vr) >> 1;
			bool ok = 1;
			vector<int> tmp;
			for (int i = 0; i < t.size(); i++) if (i != vm) tmp.pub(t[i]);
			for (int i = 0; i < tmp.size() / 2; i++) if (tmp[i] + tmp[(int)tmp.size() - i - 1] > ww){
				ok = 0;
			}
			if (ok)
				vr = vm;
			else
				vl = vm;
		}
		if (vr == t.size()){
			f = 0;
			return 0;
		} 
		return t[vr];
	}
}
 
bool can(int val){ 
	f = 1; ww = val;
	dfs(0, -1);
	return f;
}
 
int main(){
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; i++){
		int a, b;
		cin >> a >> b;
		a--; b--;
		g[a].pub(b);
		g[b].pub(a);
	}
 
	int vl = 0, vr = 1e9 + 7;
	while(vl + 1 < vr){
		int vm = (vl + vr) >> 1;
		if (can(vm))
			vr = vm;
		else
			vl = vm;
	}
	cout << calcCnt(0, -1) << ' ' << vr;
}