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
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
vector<vector<int> > ed;
vector<int> dp;
vector<int> v;
vector<int> t;
int INF = (int)1 << 30;
int N;
int m;
bool f;
int dfs(int a) {
	if (!f) {
		return -2;
	}
	v[a] = 1;
	vector<int> b;
	int k;
	for (int i = 0; i < ed[a].size(); i++) {

		if (v[ed[a][i]] == 0) {
			k = dfs(ed[a][i]);
			if (k != -1) {
				b.push_back(k);
			}
		}
		if (!f) {
			return -2;
		}
	}

	if ((int)ed[a].size() % 2 == 1) {
		b.push_back(0);
	}
	sort(b.begin(), b.end());
	
	if (b.size() % 2 == 0) {
		for (int i = 0; i < (int)b.size()/2; i++) {
			if (b[i] + b[(int)b.size() - 1 - i] > m) {
				f = false;
			}
		}
		return -1;
	}
	else {
		for (int i = 0; i < (int)(b.size() - 1) / 2; i++) {
			if (b[i] + b[(int)b.size() - 2 - i] > m) {
				f = false;
			}
		}
		int res;
		if (f) {
			res = b.back();
			int l = 0;
			for (int i = (int)b.size() - 1; i >= 1; i--) {
				if (l < i) {
					if (b[i] + b[(int)b.size() - 1 - i] > m) {
						break;
					}
				}
				else {
					if (b[i] + b[(int)b.size() - i] > m) {
						break;
					}
				}
				l++;
				res = b[i - 1];
			}
		}
		if (f) {
			return res + 1;
		}
		else {
			return -1;
		}
	}


}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;

	int res = 0;
	int a, b;
	ed.resize(N);
	for (int i = 0; i < N - 1; i++) {
		cin >> a >> b;
		a--;
		b--;
		ed[a].push_back(b);
		ed[b].push_back(a);
	}
	int u = N;
	int d = 0;
	while (u - d > 1) {
		dp.clear();
		dp.resize(N, INF);
		v.clear();
		v.resize(N, 0);
		m = (u + d) / 2;
		f = true;
		dfs(0);
		if (f) {
			u = m;
		}
		else {
			d = m;
		}
	}
	res = 0;
	for (int i = 0; i < N; i++) {
		if (ed[i].size() % 2 == 1) {
			res++;
		}
	}
	cout << (res/2) <<" " <<u << endl;
}