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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define ele int
using namespace std;
#define maxn 200010
const ele INF=1e9;
struct edge{
	ele v;
	edge *nxt;
}ep[maxn<<1],*ecnt;
ele n,r,d[maxn],f[maxn];
edge *h[maxn];
inline void addedge(ele u,ele v){
	edge *p=ecnt++;
	p->v=v; p->nxt=h[u];
	h[u]=p;
}
void dfs(ele p,ele i,ele B){
	vector<ele> lst;
	for (edge *j=h[i]; j; j=j->nxt)
		if (j->v!=p){
			dfs(i,j->v,B);
			lst.push_back(f[j->v]);
		}
	if (lst.size()%2==0) lst.push_back(0);
	sort(lst.begin(),lst.end());
	f[i]=INF;
	for (int r=0; r<lst.size(); ++r){
		ele u=0,v=lst.size()-1;
		bool flag=true;
		while (u<v){
			if (u==r) ++u;
			if (v==r) --v;
			if (u<v && lst[u]+lst[v]>B){ flag=false; break; }
			++u; --v;
		}
		if (flag){ f[i]=lst[r]+1; break; }
	}
}
inline bool test(ele B){
	dfs(-1,r,B);
	return f[r]<=B+1;
}
int main(){
	scanf("%d",&n);
	ecnt=ep; memset(h,0,sizeof(h)); memset(d,0,sizeof(d));
	for (int i=0; i<n-1; ++i){
		ele x,y;
		scanf("%d%d",&x,&y); --x,--y;
		addedge(x,y); addedge(y,x);
		++d[x]; ++d[y];
	}
	ele cnt=0;
	for (int i=0; i<n; ++i){
		if (d[i]==1) r=i;
		if (d[i]&1) ++cnt;
	}
	ele L=0,R=n-1;
	while (R-L>1){
		ele mid=(L+R)>>1;
		if (test(mid)) R=mid; else L=mid;
	}
	printf("%d %d\n",cnt/2,R);
	return 0;
}