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<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
#define pb push_back
#define sz(a) int(a.size())

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,A,lim,f[N],rt;
vector<int> E[N];
bool fl;

bool check2(vector<int> &v,int mid) {
	int l=0,r=sz(v)-1;
	while(l<r) {
		if(l==mid) { ++l;continue; }
		if(r==mid) { --r;continue; }
		if(v[l]+v[r]>lim) return 0;
		++l,--r;
	}
	return 1;
}

void dfs(int u,int fa) {
	vector<int> vec;
	for(auto v:E[u]) if(v^fa) dfs(v,u),vec.pb(f[v]);
	if(sz(vec)&1^1) vec.pb(0);
	sort(vec.begin(),vec.end());
	int l=0,r=sz(vec);
	while(l<r) {
		int mid=(l+r)>>1;
		if(check2(vec,mid)) r=mid;
		else l=mid+1;
	}
	if(l==sz(vec)) { fl=0;return; }
	f[u]=vec[l]+1;
}

bool check(int mid) {
	lim=mid;fl=1;dfs(rt,0);
	return fl&&f[rt]<=lim+1;
}

int main() {
	cin>>n;
	for(int i=1,u,v;i<n;i++) u=gi(),v=gi(),E[u].pb(v),E[v].pb(u);
	for(int i=1;i<=n;i++)
		if(sz(E[i])&1) ++A,rt=i;
	int l=0,r=n;
	while(l<r) {
		int mid=(l+r)>>1;
		if(check(mid)) r=mid;
		else l=mid+1;
	}
	cout<<A/2<<' '<<l;
	check(2);
	return 0;
}