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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int N;
vector<int> G[100005];

int dfs(int v, int prev)
{
	if(prev != -1 && G[v].size() == 1) return 1;
	
	int ret = 0;
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == prev) continue;
		ret += dfs(G[v][i], v);
	}
	if(prev == -1) ret -= (int)G[v].size() / 2;
	else ret -= ((int)G[v].size() - 2) / 2;
	return ret;
}

int dfs2(int v, int prev, int bound)
{
	if(bound <= 0) return -1;
	if(prev != -1 && G[v].size() == 1) return 1;
	
	vector<int> vec;
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == prev) continue;
		int res = dfs2(G[v][i], v, bound);
		if(res < 0) return -1;
		vec.push_back(res);
	}
	sort(vec.begin(), vec.end());
	
	if(prev == -1){
		int l = 0, r = vec.size()-1;
		bool flag = false;
		while(1){
			if(l >= r) return 1;
			if(vec[l] + vec[r] > bound) break;;
			l++, r--;
		}
	}
	else{
		if(G[v].size() % 2) vec.insert(vec.begin(), 0);
		for(int i = 0; i < vec.size(); i++){
			if(vec[i] + 1 > bound) continue;
			int l = 0, r = vec.size()-1;
			bool flag = false;
			while(1){
				if(l == i) l++;
				if(r == i) r--;
				if(l >= r) return vec[i] + 1;
				if(vec[l] + vec[r] > bound) break;
				l++, r--;
			}
		}
	}
	return -1;
}

int main(void)
{
	cin >> N;
	int a, b;
	for(int i = 1; i <= N-1; i++){
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	
	int ansA, ansB;
	ansA = dfs(1, -1);
	
	int lb = 0, ub = N, mid;
	while(ub - lb > 1){
		mid = (ub + lb) / 2;
		if(dfs2(1, -1, mid) >= 0) ub = mid;
		else lb = mid;
	}
	ansB = ub;
	
	cout << ansA << " " << ansB << endl;
	return 0;
}