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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int N; vector<int> G[100009];
int solve(int pos, int pre, int lim) {
	vector<int> v;
	for (int i : G[pos]) {
		if (i == pre) continue;
		int res = solve(i, pos, lim);
		if (res == -1) return -1;
		v.push_back(res);
	}
	if (G[pos].size() % 2 == 1) v.push_back(0);
	sort(v.begin(), v.end());
	if (pre == -1) {
		for (int i = 0; 2 * i < v.size(); i++) {
			if (v[i] + v[v.size() - i - 1] > lim) return -1;
		}
		return 0;
	}
	for (int i = 0; 2 * i < v.size() - 1; i++) {
		if (v[i] + v[v.size() - i - 2] > lim) return -1;
	}
	for (int i = v.size() - 1; i >= 1; i--) {
		int lp = v.size() - i - 1;
		if (lp >= i) lp++;
		if (v[i] + v[lp] > lim) return v[i] + 1;
	}
	return v[0] + 1;
}
int main() {
	cin >> N;
	for (int i = 1; i < N; i++) {
		int a, b;
		cin >> a >> b; a--, b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int c = 0;
	for (int i = 0; i < N; i++) c += G[i].size() % 2;
	int l = 0, r = N;
	while (r - l > 1) {
		int m = (l + r) >> 1;
		int res = solve(0, -1, m);
		if (res == -1) l = m;
		else r = m;
	}
	cout << c / 2 << ' ' << r << endl;
	return 0;
}