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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
int n,tot=0;
const int INF=1e9;
struct data {
	int next,num;
}edge[200005];
int head[200001],d[200001];
void add(int u,int v) {
	edge[++tot].next=head[u];
	edge[tot].num=v;
	head[u]=tot;
}
int dfs(int x,int fa) {
	int cnt=0;
	int sum=0;
	for (int i=head[x];i!=-1;i=edge[i].next) {
		int kx=edge[i].num;
		if (kx==fa) continue;
		cnt++;
		sum+=dfs(kx,x);
	}
	if (x==1) return sum-cnt/2;
	if (cnt%2) return sum-cnt/2;
	return sum-cnt/2+1;
}
int calc(int x,int fa,int lim) {
	vector <int> v;
	for (int i=head[x];i!=-1;i=edge[i].next) {
		int kx=edge[i].num;
		if (kx==fa) continue;
		int now=calc(kx,x,lim);
		if (now==INF) return INF;
		v.push_back(now);
	}
	int sz=v.size();
	sort(v.begin(),v.end());
	if (sz%2==0) {
		bool judge=1;
		for (int i=0;i<sz;i++)
			if (v[i]+v[sz-i-1]>lim) {
				judge=0;
				break;
			}
		if (judge) return 1;
		if (x==1) return INF;
		sz--;
	}
	int l=0,r=sz;
	while (l<r) {
		int mid=(l+r)/2;
		bool judge=1;
		vector <int> v2;
		for (int i=0;i<sz;i++) if (i!=mid) v2.push_back(v[i]);
		for (int i=0;i<sz-1;i++) if (v2[i]+v2[sz-i-2]>lim) {
			judge=0;
			break;
		}
		if (judge) r=mid;
			else l=mid+1;
	}
	if (l==sz) return INF;
	if (x==1&&v[l]>lim) return INF;
	if (x==1) return 0;
	if (v[l]+1>lim) return INF;
	return v[l]+1;
}
int work() {
	int l=1,r=2e5;
	while (l<r) {
		int mid=(l+r)/2;
		if (calc(1,-1,mid)!=INF) r=mid;
			else l=mid+1;
	}
	return l;
}
int main() {
	memset(head,-1,sizeof(head));
	scanf("%d",&n);
	for (int i=1;i<n;i++) {
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
		d[u]++;
		d[v]++;
	}
	int ans=dfs(1,-1);
	printf("%d ",ans);
	int ans2=work();
	printf("%d\n",ans2);
	return 0;
}