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 <algorithm>
#include <cstdio>
#include <vector>
#define pb push_back
std::vector<int> vec, adj[100005]; 
int dp[100005]; 
bool dfs(int u, int lim, int fa = -1)
{
	for (int v : adj[u])
	{
		if (v != fa && !dfs(v, lim, u))
			return false; 
	}
	vec.clear(); 
	for (int v : adj[u])
	{
		if (v != fa)
			vec.pb(dp[v] + 1); 
	}
	if (vec.size() & 1 ^ (fa != -1))
		vec.pb(0); 
	std::sort(vec.begin(), vec.end()); 
	if (~fa)
	{
		int l = 0, r = vec.size() - 2; 
		while (l < r)
		{
			if (vec[l++] + vec[r--] > lim)
				return false; 
		}
		int idx = vec.size() - 1; 
		while (idx && vec[idx] + vec[vec.size() - 1 - idx + (idx << 1 < vec.size())] <= lim)
			idx--; 
		dp[u] = vec[idx]; 
	}
	else
	{
		int l = 0, r = vec.size() - 1; 
		while (l < r)
		{
			if (vec[l++] + vec[r--] > lim)
				return false; 
		}
	}
	return true; 
}
int main()
{
	// freopen("ARC088-F.in", "r", stdin); 
	int n; 
	scanf("%d", &n); 
	for (int i = 1; i < n; i++)
	{
		int u, v; 
		scanf("%d%d", &u, &v); 
		adj[--u].pb(--v);
		adj[v].pb(u); 
	}
	int a = 0, l = 0, r = n - 1, ans; 
	for (int i = 0; i < n; i++)
		a += adj[i].size() & 1;
	while (l <= r)
	{
		int m = l + r >> 1; 
		if (dfs(0, m))
		{
			ans = m; 
			r = m - 1; 
		}
		else
			l = m + 1;
	}
	printf("%d %d\n", a + 1 >> 1, ans);
	return 0; 
}