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 <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;
 
#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif
 
typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair
 
const int N = 100100;
int n;
vector<int> g[N];
int dp[N];
int m;
 
bool solve(int v, int par) {
	vector<int> a;
	if ((int)g[v].size() & 1) a.push_back(0);
	for (int u : g[v]) {
		if (u == par) continue;
		if (!solve(u, v)) return false;
		if (dp[u] != -1)
			a.push_back(dp[u] + 1);
	}
	sort(a.begin(), a.end());
/*
	printf("%d :", v);
	for (int x : a)
		printf(" %d", x);
	printf("\n");
*/
	if ((int)a.size() & 1) {
		int l = -1, r = (int)a.size();
		while(r - l > 1) {
			int x = (l + r) / 2;
			int p1 = 0, p2 = (int)a.size() - 1;
			bool ok = true;
			while(p1 < p2) {
				if (p1 == x) p1++;
				if (p2 == x) p2--;
				if (a[p1] + a[p2] > m) {
					ok = false;
					break;
				}
				p1++;
				p2--;
			}
			if (ok)
				r = x;
			else
				l = x;
		}
		if (r == (int)a.size()) return false;
		dp[v] = a[r];
		return true;
	} else {
		for (int i = 0; i < (int)a.size(); i++)
			if (a[i] + a[(int)a.size() - 1 - i] > m)
				return false;
		dp[v] = -1;
		return true;
	}
}
 
int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
 
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int v, u;
		scanf("%d%d", &v, &u);
		v--;u--;
		g[v].push_back(u);
		g[u].push_back(v);
	}
	int ans = 0;
	for (int i = 0; i < n; i++)
		ans += (int)g[i].size() & 1;
	printf("%d ", ans / 2);
	int l = 0, r = n;
	while(r - l > 1) {
		m = (l + r) / 2;
		if (solve(0, -1))
			r = m;
		else
			l = m;
	}
	printf("%d\n", r);
 
	return 0;
}