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 maxn = 100100;
vector<int> to[maxn];
int lim;
bool can;
int solve(int cur, int par){
	vector<int> f;
	for(int nxt : to[cur]){
		if(nxt == par) continue;
		f.push_back(solve(nxt, cur));
	}
	if(f.size() % 2 == 0) f.push_back(0);
	sort(f.begin(), f.end());
	int len = f.size();
	int lo = 0, hi = len - 1, ans = -1;
	while(lo <= hi){
		int mid = (lo + hi)>>1;
		auto get = [&](int x){
			if(x >= mid) x++;
			return f[x];
		};
		bool ok = true;
		for(int i = 0; i < (len - 1)/2; i++){
			int ri = (len - 2) - i;
			ok &= get(i) + get(ri) <= lim;
		}
		if(ok) ans = mid, hi = mid - 1;
		else lo = mid + 1;
	}
	if(ans == -1){
		can = false;
		return 0;
	}
	return f[ans] + 1;
}
int main(){
	int n; cin >> n;
	for(int i = 0; i < n - 1; i++){
		int a, b; cin >> a >> b; a--; b--;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	int A = 0;
	for(int i = 0; i < n; i++) A += to[i].size()%2;
	int leaf = -1;
	for(int i = 0; i < n; i++) if(to[i].size() == 1) leaf = i;
	assert(~leaf);
	int lo = 1, hi = n - 1, ans = n;
	while(lo <= hi){
		int mid = (lo + hi)>>1;
		lim = mid;
		can = true;
		if(solve(leaf, -1) <= lim + 1 && can){
			ans = mid;
			hi = mid - 1;
		} else {
			lo = mid + 1;
		}
	}
	cout << A/2 << " " << ans << endl;
	return 0;
}