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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int MAX_N=1e5,INF=1e8;
bool ans;
int B,dp[MAX_N],N;
vector<int> G[MAX_N];
void dfs(int v,int p){
	vector<int> vec;
	int n=0;
	for(auto e:G[v]){
		if (e!=p){
			dfs(e,v);
			vec.push_back(dp[e]);
			n++;
		}
	}
	sort(vec.begin(),vec.end());
	dp[v]=INF;
	if (n%2==0){
		bool f=true;
		REP(i,n/2){
			if (vec[i]+vec[n-i-1]+2>B){
				f=false;
			}
		}
		if (f){
			dp[v]=0;
		}
		if (n==0){
			return ;
		}
	}
	if (p!=-1 || n%2){
		if (n%2==0){
			if (vec[n-1]+1>B){
				return;
			}
			n--;
		}
		if (vec[n-1]+1>B){
			return;
		}
		int lb=-1,ub=n;
		while(ub-lb>1){
			int mid=(lb+ub)/2;
			int le=0,ri=n-1;
			bool f=true;
			REP(i,n/2){
				if (le==mid){
					le++;
				}
				if (ri==mid){
					ri--;
				}
				if (vec[le]+vec[ri]+2>B){
					f=false;
				}
				le++;
				ri--;
			}
			if (f){
				ub=mid;
			}else{
				lb=mid;
			}
		}
		if (ub!=n){
			dp[v]=min(dp[v],vec[ub]+1);
		}
	}
}

int main(){
	cin>>N;
	REP(i,N-1){
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int A=0;
	REP(i,N){
		if (G[i].size()%2){
			A++;
		}
	}
	A/=2;
	int lb=0,ub=N-1;
	while(ub-lb>1){
		int mid=(lb+ub)/2;
		B=mid;
		ans=true;
		dfs(0,-1);
		REP(i,N){
			if (dp[i]==INF){
				ans=false;
			}
		}
		if (ans){
			ub=mid;
		}else{
			lb=mid;
		}
	}
	cout<<A<<" "<<ub<<endl;
	return 0;
}