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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>pat[101000];
vector<int>ko[101000];
bool flag[101000];
int deg[101000];
void dfs(int node)
{
	flag[node] = true;
	for (int i = 0; i < pat[node].size(); i++)
	{
		if (!flag[pat[node][i]])
		{
			dfs(pat[node][i]);
			ko[node].push_back(pat[node][i]);
		}
	}
}
int calc(int node, int med)
{
	vector<int>vec;
	for (int i = 0; i < ko[node].size(); i++)vec.push_back(calc(ko[node][i], med));
	sort(vec.begin(), vec.end());
	if (vec.size() % 2 == 0)
	{
		bool f = true;
		for (int i = 0; i < vec.size() / 2; i++)if (vec[i] + vec[vec.size() - 1 - i]>med)f = false;
		if (f)return 1;
		if (vec[vec.size() - 1]>med)return 10000000;
		vec.pop_back();
		int b = 0, e = vec.size();
		for (;;)
		{
			if (b == e)break;
			int m = (b + e) / 2;
			vector<int>z;
			for (int i = 0; i < vec.size(); i++)if (m != i)z.push_back(vec[i]);
			bool g = true;
			for (int i = 0; i < z.size() / 2; i++)if (z[i] + z[z.size() - 1 - i]>med)g = false;
			if (g)e = m;
			else b = m + 1;
		}
		if (b == vec.size())return 10000000;
		else return vec[b] + 1;
	}
	else
	{
		int b = 0, e = vec.size();
		for (;;)
		{
			if (b == e)break;
			int m = (b + e) / 2;
			vector<int>z;
			for (int i = 0; i < vec.size(); i++)if (m != i)z.push_back(vec[i]);
			bool g = true;
			for (int i = 0; i < z.size() / 2; i++)if (z[i] + z[z.size() - 1 - i]>med)g = false;
			if (g)e = m;
			else b = m + 1;
		}
		if (b == vec.size())return 10000000;
		else return vec[b] + 1;
	}
}
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)pat[i].clear(), ko[i].clear(), flag[i] = false, deg[i] = 0;
	for (int i = 0; i < num - 1; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		za--, zb--;
		pat[za].push_back(zb);
		pat[zb].push_back(za);
		deg[za]++, deg[zb]++;
	}
	int root;
	for (int i = 0; i < num; i++)if (deg[i] == 1)root = i;
	dfs(root);
	int beg = 1, end = num - 1;
	for (;;)
	{
		if (beg == end)break;
		int med = (beg + end) / 2;
		if (calc(root, med) <= med + 1)end = med;
		else beg = med + 1;
	}
	int r = 0;
	for (int i = 0; i < num; i++)r += deg[i] % 2;
	printf("%d %d\n", r / 2, beg);
}