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 <bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 5;
 
int n;
vector<int> g[N];
int A, newA, lim;
 
int dfs(int u,int p) {
	multiset<int> ms;
	for (int v : g[u]) if (v != p) {
		ms.insert(dfs(v, u) + 1);
	}
	
	vector<int> rem;
	while (!ms.empty()) {
		multiset<int>::iterator it = ms.end();
		int x = *(--it);
		ms.erase(it);
		it = ms.upper_bound(lim - x);
		if (it == ms.begin()) rem.push_back(x);
		else ms.erase(--it), newA++; 
	}
	
	reverse(rem.begin(), rem.end());
	while (!rem.empty() && rem.back() == lim) newA++, rem.pop_back();
	if (rem.empty()) return 0;
	else { newA += rem.size() - 1; return rem[0]; }
}
 
bool check() {
	newA = 0;
	newA += dfs(1, 1) > 0;
	return newA == A;
}
 
int main() {
	ios_base::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n - 1; ++i) {
		int u, v; cin >> u >> v;
		g[u].push_back(v), g[v].push_back(u);
	}
	for (int i = 1; i <= n; ++i) {
		A += g[i].size() & 1;
	}
	A /= 2;
	int l = 1, r = n;
	while (l < r) {
		lim = l + r >> 1;
		if (check()) r = lim;
		else l = lim + 1;
	}
	cout << A << ' ' << l << '\n';
}