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 long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 100010;
int N;
vector<int> V[MC];
vector<int> S[MC];
int r;

bool dfs(int p , int q , int z){
	S[q].clear();
	for(auto u : V[q]){
		if(u==p) continue;
		if(!dfs(q,u,z)) return 0;
	}
	if(S[q].size()%2==0) S[q].PB(0);
	sort(S[q].begin(),S[q].end());
	int f = 0 , g = -1;
	deque<int> Q;
	repm(i,S[q].size()-1,-1){
		while(f<i && S[q][f]+S[q][i]<=z){
			Q.push_back(S[q][f]);
			++f;
		}
		if(f == i){
			Q.push_back(S[q][f]);
			break;
		}
		if(Q.empty()){
			if(~g) return 0;
			g = S[q][i];
		} else {
			Q.pop_back();
		}
	}
	int w = max(Q.empty()?0:Q.front(),g);
	if(p>0) S[p].PB(w+1);
	else if(w>z) return 0;
	return 1;
}

int BS(int x , int y){
	if(x-y<2) return x;
	int z = (x+y) / 2;
	return dfs(-1,r,z) ? BS(z,y) : BS(x,z);
}

int main(){
	cin >> N;
	repp(i,1,N){
		int a,b;
		cin >> a >> b;
		V[a].PB(b);
		V[b].PB(a);
	}
	int A = 1;
	repp(i,1,N+1){
		A += (V[i].size()-1)/2;
		if(V[i].size()==1) r = i;
	}
	cout << A << ' ' << (A==1?N-1:BS(N-1,0)) << endl;
	return 0;
}
