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;
const int maxn = 1e5 + 10;
int n, A, B, tot;
vector<int> E[maxn];
int dfs(int x, int fa)
{
	int ret = 0;
	for(auto v : E[x])
		if(v != fa)
			ret += dfs(v, x);
	if(fa) ret += (E[x].size() - 1) / 2;
	else ret += (E[x].size() + 1) / 2;
	return ret;
}
int color(int x, int fa)
{
	vector<int> a;
	for(auto v : E[x])
		if(v != fa)
		{
			int res = color(v, x);
			if(res == -1) return -1;
			else if(res == B) ++tot, res = 0;
			a.push_back(res + 1);
		}
	sort(a.begin(), a.end());
	if(!fa)
	{
		if(a.size() % 2 == 0)
		{
			for(int i = 0, j = a.size() - 1; i < j; ++i, --j)
				if(a[i] + a[j] > B)
					return -1;
			tot += a.size() / 2;
			return 0;
		}
		else
		{
			for(int i = 0, j = a.size() - 2; i < j; ++i, --j)
				if(a[i] + a[j] > B)
					return -1;
			tot += a.size() / 2 + 1;
			return a.back();
		}
	}
	else
	{
		if(a.size() % 2 == 0)
		{
			bool flag = 1;
			for(int i = 0, j = a.size() - 1; i < j; ++i, --j)
				if(a[i] + a[j] > B)
				{
					flag = 0;
					break;
				}
			if(flag) 
			{
				tot += a.size() / 2;
				return 0;
			}
			a.pop_back();
			++tot;
		}
		int ret = a[0];
		for(int i = 1, j = a.size() - 1; i < j; ++i, --j)
			if(a[i] + a[j] > B)
			{
				if(ret + a[j] <= B) ret = a[i];
				else
				{
					ret = a[j];                                                                                                  
					a.erase(a.begin() + j);
					for(int ii = 0, jj = a.size() - 1; ii < jj; ++ii, --jj)
						if(a[ii] + a[jj] > B)
							return -1;
					tot += a.size() / 2;
					return ret;
				}
			}
		tot += a.size() / 2;
		return ret;
	}
}
bool check()
{
	tot = 0;
	if(color(1, 0) == -1) return 0;
	return tot <= A;
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i < n; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	A = dfs(1, 0);
	int l = 1, r = n - 1;
	while(l <= r)
	{
		B = l + r >> 1;
		if(check()) r = B - 1;
		else l = B + 1;
	}
	cout << A << ' ' << l << endl;
	return 0;
}