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 <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 100010 * 2;
const int INF = 1<<30;

vector<int> g[maxn];
int father[maxn];
int root = 1;
bool flg;
int dp[maxn];
int n;

void link(int u, int v)
{
	g[u].push_back(v);
	g[v].push_back(u);
}

bool check2(int r, vector<int> son, int B)
{
	son.erase(son.begin()+r);
	for (int i=0;i<son.size()/2;i++)
	{
		int j = son.size() - 1 - i;
		if (son[i] + son[j] > B) return false;
	}
	return true;
}

void dfs2(int u, int f, int B)
{
	vector<int> son; son.clear();
	for (int i=0;i<g[u].size();i++)
	{
		int v = g[u][i];
		if (v == f) continue;
		dfs2(v,u,B);
		son.push_back(dp[v]);
	}
	sort(son.begin(), son.end());
	int size = son.size();

	if (size == 0)
	{
		dp[u] = 1;
	}
	else if (size % 2 == 1) //u为偶度点
	{
		int lb = 0, ub = size-1; //[lb,ub]
		while (ub - lb >= 1)
		{
			int mid = (ub + lb)/2;
			if (check2(mid,son,B)) ub = mid;
			else lb = mid+1;
		}
		if (!check2(lb,son,B)) flg = false; //无论怎么样都满足不了要求
		dp[u] = son[lb] + 1;
		if (dp[u] > B && u != root) flg = false; 
	}
	else //u为奇度点
	{
		son.push_back(0); //如果选中0表示u作为新的起点，如果没有选中，表示有一个儿子是通过u出去的
		sort(son.begin(), son.end());
		int lb = 0, ub = size; //[lb,ub]
		while (ub - lb >= 1)
		{
			int mid = (ub + lb)/2;
			if (check2(mid,son,B)) ub = mid;
			else lb = mid+1;
		}
		dp[u] = son[lb] + 1;
		if (!check2(lb,son,B)) flg = false;
		if (dp[u] > B && u != root) flg = false; 
	}
}

bool check1(int B)
{
	for (int i=0;i<=n+10;i++) dp[i] = INF;
	flg = true;
	dfs2(root,0,B);
	//cout<<B<<" "<<flg<<" "<<dp[root]<<endl;
	if (flg == false || dp[root] > B+1) return false;
	else return true;
}

int main()
{
	//freopen("F.in","r",stdin);
	scanf("%d",&n);
	for (int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		link(u,v);
	}
	for (int i=1;i<=n;i++)
	{
		if (g[i].size() == 1)
		{
			root = i;
			break;
		}
	}

	int oddNum = 0;
	for (int i=1;i<=n;i++) oddNum += g[i].size() % 2;
	int A = oddNum / 2;

	int Blb = 0, Bub = n-1; //[Blb,Bub]
	while (Bub - Blb >= 1)
	{
		int mid = (Blb + Bub)/2;
		if (check1(mid)) Bub = mid;
		else Blb = mid+1;
	}

	printf("%d %d\n",A,Bub);
	return 0;
}