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>
#define MN 100005
using namespace std;
inline int in(){
    int x=0;bool f=0; char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
struct st{
	int to,nxt;
}e[MN<<1];
int head[MN],a[MN],st[MN],f[MN],ind[MN],fa[MN];
int cnt=0,num=0,top,n,x,y,rt,ans;
bool mk;
inline void ins(int x,int y){
	e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;
}
inline bool chk(int u,int x,int tp){
	int hd=1,tl=tp;
	if (u==hd) ++hd;if (u==tl) --tl;
	while (hd<tl){
		if (st[hd]+st[tl]>x) return 0;++hd;--tl;
		if (u==hd) ++hd;if (u==tl) --tl;
	}return 1;
}
inline void dfs(int u,int x){
	for (int i=head[u];i;i=e[i].nxt){
		int v=e[i].to;
		if (v!=fa[u]) fa[v]=u,dfs(v,x);
	}top=0;
	for (int i=head[u];i;i=e[i].nxt){
		int v=e[i].to;
		if (v!=fa[u]) st[++top]=f[v];
	}if (!top) {f[u]=1;return;}
	if (!(top&1)) st[++top]=0;
	sort(st+1,st+top+1);int l=1,r=top,res=-1;
	while (l<=r){
		int mid=(l+r)>>1;
		if (chk(mid,x,top)) res=mid,r=mid-1;
		else l=mid+1;
	}if (res==-1&&!chk(r,x,top)) mk=0;f[u]=st[res]+1;	
}
inline bool check(int x){
	mk=1;fa[rt]=0;dfs(rt,x);
	return (mk&&(f[rt]<=x+1));	
}
int main()
{
	n=in();
	for (int i=1;i<n;++i){
		x=in();y=in();ins(x,y);ins(y,x);
		++ind[x];++ind[y];
	}
	for (int i=1;i<=n;++i) if (ind[i]&1) rt=i,++num;
	printf("%d ",num>>1);
	int l=1,r=n;while (l<=r){
		int mid=(l+r)>>1;
		if (check(mid)) ans=mid,r=mid-1;
		else l=mid+1;
	}printf("%d",ans);return 0;
}