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+10;
int n,res[N],A,B,l,r,mid;
vector<int> e[N];
bool ok;
int rest(vector<int> &L){
	if (!(L.size()&1)) L.push_back(0);
	sort(L.begin(),L.end());
	for (int l=0,r=L.size()-2;l<r;l++,r--)
	if (L[l]+L[r]>mid) ok=0;
	int p=L.size()-1,len=p;
	while (p&&L[p]+L[len-p+(len-p>=p)]<=mid) p--;
	return L[p];
}
void solve(int x,int fa){
	vector<int> L;
	for (auto v:e[x])
	if (v!=fa) solve(v,x),L.push_back(res[v]+1);
	if (fa) res[x]=rest(L);
	else{
		if (L.size()&1) L.push_back(0);
		sort(L.begin(),L.end());
		for (int l=0,r=L.size()-1;l<r;l++,r--)
		if (L[l]+L[r]>mid) ok=0;
	}
}
bool check(){
	ok=1;
	solve(1,0);
	return ok&&res[1]<=mid;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	A=n-1;
	for (int i=1;i<=n;i++) A-=e[i].size()/2;
	for (l=0,r=n-1;l<r;){
		mid=(l+r)>>1;
		if (check()) r=mid;else l=mid+1;
	}
	printf("%d %d\n",A,l);
	return 0;
}