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>
using namespace std;
const int N=100005;
int n,ed,B,root,g[N],f[N],q[N],deg[N];
struct E{int v,nxt;}e[N<<1];
inline void adde(int x,int y){e[++ed].v=y;e[ed].nxt=g[x];g[x]=ed;}
int dfs(int x,int y){
	int i;f[x]=0;
	for(i=g[x];i;i=e[i].nxt)if(e[i].v!=y&&!dfs(e[i].v,x))return 0;
	int t=0;
	for(i=g[x];i;i=e[i].nxt)if(e[i].v!=y)q[++t]=f[e[i].v]+1;
	if(!t)return 1;
	if(~t&1)q[++t]=0;
	sort(q+1,q+1+t);
	if(q[t]>B)return 0;
	int l=1,r=t+1;
	for(;l<r;){
		int mid=(l+r)>>1;
		int l1=0,r1=t+1;
		for(i=1;i<=t/2;++i){
			if(++l1,l1==mid)++l1;
			if(--r1,r1==mid)--r1;
			if(q[l1]+q[r1]>B)break;
		}
		if(i<=t/2)l=mid+1;
		else r=mid;
	}
	if(l>t)return 0;
	return f[x]=q[l],1; 
}

int main(){
	//freopen("aa.in","r",stdin);
	int i,j;
	scanf("%d",&n);
	for(i=1;i<n;++i){
		int x,y;scanf("%d%d",&x,&y);
		adde(x,y);adde(y,x);
		++deg[x];++deg[y];
	}
	int tot=0;
	for(i=1;i<=n;++i){
		if(deg[i]&1)++tot;
		if(deg[i]==1)root=i;	
	}
	tot>>=1;
	int l=1,r=n;
	for(;l<r;){
		int mid=(l+r)>>1;B=mid;
		if(!dfs(root,root))l=mid+1;
		else r=mid;
	}
	printf("%d %d",tot,l);
	return 0;
}