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<algorithm>
#include<cstring>
#include<set>
#define MN 100005
using namespace std;
int n,num,cnt,now[MN],f[MN],head[MN],pre[MN],lst[MN];
struct edge{int to,next;}g[MN<<1];
multiset<int>::iterator it,it2;
void ins(int u,int v){g[++num].next=head[u];head[u]=num;g[num].to=v;}
bool dp(int u,int ff,int mid){
	int tt=0,lef=0;
	for(int i=head[u];i;i=g[i].next)if(g[i].to!=ff)if(!dp(g[i].to,u,mid))return false;
	multiset<int>S;S.clear();
	for(int i=head[u];i;i=g[i].next)if(g[i].to!=ff)S.insert(f[g[i].to]),tt++;
	if(!tt){f[u]=0;return true;}
	while(S.begin()!=S.end()){
		it=S.end();--it;int tmp=(*it);S.erase(it);
		it2=S.upper_bound(mid-tmp-2);
		if(it2!=S.begin()){--it2;S.erase(it2);}
		else {lst[++lef]=tmp;}
	}
	if(u==1){
		if(lef==1)return lst[1]+1<=mid;
		else return lef==0;
	}else {
		if(tt&1){
			if(lef==1){f[u]=lst[lef]+1;return lst[1]+1<=mid;}
			else {f[u]=0;return lef==0;}
		}else {
			if(lef==2){f[u]=lst[2]+1;return f[u]<=mid&&lst[1]+1<=mid;}
			else {f[u]=0;return lef==0;}
		}
	}
}
void calc(int u,int ff){
	int tt=0;
	for(int i=head[u];i;i=g[i].next)if(g[i].to!=ff)calc(g[i].to,u),tt++;
	cnt+=(u==1?(tt+1)/2:tt/2);
}
int main(){
	scanf("%d",&n);int x,y;
	for(int i=1;i<n;i++)scanf("%d%d",&x,&y),ins(x,y),ins(y,x);
	int l=1,r=1e5,ans=1e5;
	while(l<=r){
		int mid=(l+r)>>1;
		if(dp(1,1,mid))ans=mid,r=mid-1;
		else l=mid+1;
	}calc(1,1);printf("%d %d",cnt,ans);
}