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 <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
const int MN = 100100;
vi g[MN];
int dp[MN];
int rt,b;
int ds(int u, int p) {
	if(dp[u] != -1) return dp[u];
	vi w;
	for(int i=0;i<g[u].size();i++) {
		int v = g[u][i];
		if(v == p) continue;
		w.push_back(ds(v,u));
	}
	if(!(w.size()&1)) {w.push_back(0);}
	sort(w.begin(),w.end());
	int s = 0,e = w.size()-1;
	int ls = -1;
	while(s <= e) {
		int st = 0,ed = w.size()-1;
		int m = (s+e)/2;
		bool pp = true;
		while(1) {
			if(st == m) {st++;}
			if(ed == m) {ed--;}
			if(st > ed) break;
			if(w[st]+w[ed] > b) {
				pp = false;break;
			}
			st++;ed--;
		}
		if(pp) {
			ls = m;e = m-1;
		} else {
			s = m+1;
		}
	}
	if(ls < 0) {dp[u] = MN*9;} else {
		dp[u] = w[ls]+1;
	}
	return dp[u];
}
int main() {
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;
	cin >> n;
	for(int i=0;i<n-1;i++) {
		int a,b;
		cin >> a >> b;a--;b--;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	int no = 0;
	for(int i=0;i<n;i++) {
		if(g[i].size() == 1) {rt = i;}
		if(g[i].size()&1) {no++;}
	}
	int s = 1,e = n;
	int ls = n;
	while(s <= e) {
		memset(dp,-1,sizeof(dp));
		int m = (s+e)/2;
		b = m;
		ds(rt,-1);
		bool po = true;
		for(int i=0;i<n;i++) {
			if(dp[i] > b + (i==rt?1:0)) {
				po = false;break;
			}
		}
		if(po) {
			ls = m;e = m-1;
		} else {
			s = m+1;
		}
	}
	cout << no/2 << " " << ls << '\n';
}