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<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN = 100000+9;
struct E{
	int next,to;
}edge[MAXN<<1];
int head[MAXN],edge_num;
int n;
int degree[MAXN];
int dp[MAXN];
int A,B;
int limit;

void addedge(int x,int y){
	degree[x]++;
	edge[++edge_num].next=head[x];
	edge[edge_num].to=y;
	head[x]=edge_num;
}

bool DP(int x,int fa){
	vector<int> vec;
	int cnt=0;
	if(degree[x]%2==1)
		vec.push_back(0);
	for(int i=head[x];i;i=edge[i].next){
		if(edge[i].to!=fa){
			if(!DP(edge[i].to,x))return false;
			if(dp[edge[i].to]!=-1)vec.push_back(dp[edge[i].to]+1);
		}
	}

	sort(vec.begin(),vec.end());
	if(vec.size()%2==1){
		int loc=-1;
		int l=0,r=vec.size()-1;
		while(l<=r){
			int mid=l+r>>1;
			bool jud=1;
			int p1=0,p2=vec.size()-1;
			while(p1<p2){
				if(p1==mid)p1++;
				if(p2==mid)p2--;
				if(vec[p1]+vec[p2]>limit){
					jud=0;
					break;
				}
				p1++;p2--;
			}
			if(jud){
				loc=mid;
				r=mid-1;
			}
			else
				l=mid+1;
		}
		if(loc==-1){
			return false;
		}
		dp[x]=vec[loc];
		return true;
	}
	else{
		for(int i=0,j=vec.size()-1;i<j;i++,j--){
			if(vec[i]+vec[j]>limit){
				return false;
			}
		}
		dp[x]=-1;return true;
	}
}

bool check(int len){
	limit=len;
	memset(dp,0,sizeof(dp));
	return DP(1,0);
}

void solve(){
	int l=1,r=n;
	while(l<=r){
		int mid=l+r>>1;
		if(check(mid)){
			r=mid-1;
			B=mid;
		}
		else
			l=mid+1;
	}
}

int main(){
	//freopen("in.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		addedge(a,b);
		addedge(b,a);
	}
	for(int i=1;i<=n;i++){
		if(degree[i]%2==1)
			A++;
	}
	A/=2;
	solve();
	printf("%d %d\n",A,B);
	return 0;
}