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
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#  include <intrin.h>
#  define __builtin_popcount __popcnt
#endif

#define _USE_MATH_DEFINES

#include <stdio.h>
#include <iostream>
#include <vector>
#include <unordered_map>
#include <stack>
#include <list>
#include <math.h>
#include <map>
#include <iomanip>
#include <functional>
#include <string>
#include <algorithm>
#include <queue>
#include <set>

using namespace std;

template<class T> ostream& operator<<(ostream& os, const vector<T>& v) {
	for (auto i = begin(v); i != end(v); i++) os << *i << (i == end(v) - 1 ? "" : " "); return os;
}
template<class T> istream& operator>>(istream& is, vector<T>& v) {
	for (auto i = begin(v); i != end(v); i++) is >> *i; return is;
}

vector<int> edg[101000];

int dfs(int i, int max_sz, int parent = -1)
{
	vector<int> arr;
	for (int u : edg[i])
	{
		if (u == parent)
			continue;
		int add = dfs(u, max_sz, i) + 1;
		arr.push_back(add);
		if (add > max_sz)
			throw 1;
	}

	
	if (arr.size() % 2 == 0)
	{
#if 0
		bool ok = true;
		for (int i = 0, j = arr.size() - 1; i < j; i++, j--)
		{
			if ((arr[i] + arr[j]) > max_sz)
				ok = false;
		}
		if (ok)
			return 0;
		arr.pop_back();
#endif
		arr.push_back(0);
	}

	sort(arr.begin(), arr.end());

	{
		int left = 0, right = arr.size();
		while (left < right)
		{
			int m = (left + right) / 2;

			vector<int> tmp;
			for (int j = 0; j < arr.size(); j++)
			{
				if (j != m)
					tmp.push_back(arr[j]);
			}

			bool ok = true;
			for (int i = 0, j = tmp.size() - 1; i < j; i++, j--)
			{
				if ((tmp[i] + tmp[j]) > max_sz)
					ok = false;
			}

			if (ok)
				right = m;
			else
				left = m + 1;
		}

		if (left == arr.size())
			throw 1;
		return arr[left];
	}
}

int main()
{
#if defined(_DEBUG) 
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	cin.tie(NULL);
	ios_base::sync_with_stdio(false);
	cout.tie(NULL);

	int n;
	cin >> n;
	for (int i = 0; i + 1 < n; i++)
	{
		int a, b;
		cin >> a >> b;
		a--, b--;
		edg[a].push_back(b);
		edg[b].push_back(a);
	}

	int A = 0;
	int leaf_i = 0;
	for (int i = 0; i < n; i++)
	{
		if (edg[i].size() % 2 == 1)
			A++;
		if (edg[i].size() == 1)
			leaf_i = i;
	}

	A /= 2;

	int left = 0, right = n;
	while (left < right)
	{
		int m = (left + right) / 2;
		try
		{
			dfs(leaf_i, m, -1);
		}
		catch (int)
		{
			left = m + 1;
			continue;
		}
		right = m;
	}

	cout << A << " " << left;

	return 0;
}