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 inf 233366666
using namespace std;
inline int read()
{int x=0;
char c=getchar();
while (c<'0'||c>'9') c=getchar();
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x;
}
int deg[100005],head[100005],nxt[200005],v[200005];
int dp[100005],n,tot=0,root=-1,tar;
inline void add(int a,int b)
{tot++;nxt[tot]=head[a];head[a]=tot;v[tot]=b;}
void dfs(int pos,int f)
{int l,r,i,ts,num=0;
for (i=head[pos];i;i=nxt[i])
{if (v[i]==f) continue;
dfs(v[i],pos);num++;
}
int tp[num+2],sz=0;
for (i=head[pos];i;i=nxt[i])
{if (v[i]==f) continue;
tp[++sz]=dp[v[i]];
}
if (!(sz&1)) tp[++sz]=0;
sort(tp+1,tp+sz+1);
l=1;r=sz;ts=sz>>1;
while (l<=r)
{int mid=(l+r)>>1,p1=1,p2=sz,tag=1;
for (i=1;i<=ts;i++)
{if (p1==mid) p1++;
if (p2==mid) p2--;
if (tp[p1]+tp[p2]>tar) {tag=0;break;}
p1++;p2--;
}
if (tag) {r=mid-1;}
else {l=mid+1;}
}
if (l==sz+1) {dp[pos]=inf;}
else {dp[pos]=tp[l]+1;}
}
inline bool chk(int mid)
{tar=mid;dfs(root,0);
return (dp[root]<=mid+1);
}
int main (){
	int i,a,b,cnt=0;
	n=read();
	for (i=1;i<n;i++)
	{a=read();b=read();
	add(a,b);add(b,a);
	deg[a]++;deg[b]++;
	}
	for (i=1;i<=n;i++)
	{if (deg[i]&1)
	{cnt++;root=i;}
	}
	cnt>>=1;
	int l=1,r=n;
	while (l<=r)
	{int mid=(l+r)>>1;
	if (chk(mid)) {r=mid-1;}
	else {l=mid+1;}
	}
	printf ("%d %d\n",cnt,l);
	return 0;
}
	
	