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<bits/stdc++.h>
#define N 100005
#define M 200005
#define pb push_back
using namespace std;
int i,j,k,l,s,n,m,last[N],to[M],Next[M],q[N],fa[N],si[N],ans,tot,x,y,dis[N],a[N];
vector<int>GT;
inline void add(int x,int y) {
	Next[++tot]=last[x]; last[x]=tot; to[tot]=y;
}
inline void build() {
	int l=0,r=1; q[1]=1,dis[1]=1;
	while (l<r) {
		int k=q[++l];
		for (int i=last[k];i;i=Next[i]) if (fa[k]!=to[i]) fa[q[++r]=to[i]]=k,dis[to[i]]=dis[k]+1;
	}
}
inline bool work(int x,int y,int z) {
	int L=0,R=GT.size()-1;
	if (L==x) L++;
	if (R==x) R--;
	while (L<=R) {	
		if (GT[L]+GT[R]-2*y+1>z) return false;
		L++,R--;
		if (L==x) L++;
		if (R==x) R--;
	}
	return true;
}
inline int Find(int x,int y) {
	int L=0,R=GT.size()-1,s=GT.size();
	while (L<=R) {
		int mid=(L+R)>>1;
		if (work(mid,x,y)) s=mid,R=mid-1;
		else L=mid+1;
	}
	return s;
}
inline bool check(int x) {
	for (int i=n;i;i--) {
		GT.clear();
		for (int j=last[q[i]];j;j=Next[j]) {
			if (to[j]==fa[q[i]]) continue;
			GT.pb(a[to[j]]);
		}
		sort(GT.begin(),GT.end());
		if (!GT.size()) {
			a[q[i]]=dis[q[i]];
			continue;
		}
 		if (GT.size()&1) {
			if (GT.size()==1) {
				a[q[i]]=GT[0];
				continue;
			}
			int gt=Find(dis[q[i]],x);
			if (gt>=GT.size()) return false;
			a[q[i]]=GT[gt];
		}
		else if (i!=1) {
			GT.pb(dis[q[i]]);
			sort(GT.begin(),GT.end());
			int gt=Find(dis[q[i]],x);
			if (gt>=GT.size()) return false;
			a[q[i]]=GT[gt];
		}
		else {
			for (int j=0;j<GT.size()/2;j++) if (GT[j]+GT[GT.size()-j-1]-1>x) return false;
			return true;
		}
	}
	if (a[1]>x) return false;
	return true;
}
int main() {
	scanf("%d",&n);
	for (i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	build();
	for (i=n;i;i--) {
		if (!si[q[i]]) si[q[i]]++;
		ans+=si[q[i]]>>1;
		si[fa[q[i]]]++;
	}
	ans+=si[1]&1;
	printf("%d ",ans);
	int l=0,r=n;
	ans=0;
	while (l<=r) {
		int mid=(l+r)>>1;
		if (check(mid)) r=mid-1,ans=mid;
		else l=mid+1;
	}
	printf("%d\n",ans-1);
} 