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 <cstdio>
#include <vector>
#include <algorithm>
#include <cassert>

using namespace std;

int B, r;
vector<int> g[100001];

int f(int x, int y) {
	vector<int> v;
	for (auto t : g[x]) if (t != y) {
		int z = f(t, x) + 1;
		if (!z) return -1;
		if (z == B) r++;
		else v.push_back(z);
	}
	int i, j, t = 0;
	sort(v.begin(), v.end());
	for (i = 0, j = (int)v.size() - 1; i < j; j--) {
		if (v[i] + v[j] <= B) r++, i++;
		else {
			if (t) r++;
			t = v[j];
		}
	}
	if (i == j) {
		if (t) r++;
		for (j++; j < v.size() && v[i] + v[j] <= B; i--, j++);
		assert(i >= 0);
		return v[i];
	}
	return t;
}

int main() {
	int L, R;
	int i, j, k, n, t, tr;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d", &j, &k);
		g[j].push_back(k);
		g[k].push_back(j);
	}
	tr = 1;
	for (i = 1; i <= n; i++) tr += (g[i].size() - 1) / 2;
	B = n;
	t = f(1, 1);
	if (t) r++;
	assert(r == tr);
	L = 1;
	R = n;
	while (L < R) {
		B = (L + R) / 2;
		r = 0;
		t = f(1, 1);
		if (t) r++;
		if (r == tr) R = B;
		else L = B + 1;
	}
	printf("%d %d\n", tr, L);
	return 0;
}