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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N;
vector<vector<int>> v(100010);
int visited[100010] = {0},dp[100010] = {0};

bool ok(int n,int m){
	visited[n] = 1;
	bool judge = true;
	vector<int> children;
	for(int i=0;i<v[n].size();i++){
		int s = v[n][i];
		if(visited[s]==0){
			judge = (judge && ok(s,m));
			children.push_back(dp[s]);
		}
	}
	int len = children.size();
	sort(children.begin(),children.end());
	if(len%2==0){
		bool j = true;
		for(int i=0;i<len/2;i++){
			if(children[i]+children[len-i-1]>m) j = false;
		}
		if(j) return judge;
		else if(n==1) return false;
		children.pop_back(); len--;
	}
	int l = -1,r = len;
	while(l+1<r){
		int mid = (l+r)/2;
		int a = 0,b = len-1;
		bool j = true;
		while(a<b){
			if(a==mid) a++;
			if(b==mid) b--;
			if(children[a]+children[b]>m) j = false;
			a++; b--;
		}
		if(j && children[mid]+dp[n]<=m) r = mid;
		else l = mid;
	}
	if(r==len) judge = false;
	else dp[n] += children[r];
	if(n==1 && dp[n]>m) judge = false;
	return judge;
}

int main(){
	cin >> N;
	int a,b;
	for(int i=0;i<N-1;i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	int A = 0,cnt = 0;
	for(int i=1;i<=N;i++){
		if(v[i].size()%2==1) cnt++;
	}
	A = cnt/2;
	int l = 0,r = N;
	while(l+1<r){
		dp[1] = 0;
		for(int i=2;i<=N;i++){
			visited[i] = 0; dp[i] = 1;
		}
		int m = (l+r)/2;
		if(ok(1,m)) r = m;
		else l = m;
	}
	cout << A << " " << r << endl;
}