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 <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> G[100100];
int N,Y,U,A;

int dfs(int x, int l, int m)
{
	vector<int> u;
	for (auto &y : G[x]) if (y != l){
		u.push_back(dfs(y,x,m)+1);
	}
	sort(u.begin(),u.end());
	int ret = 0;
	if (u.size() % 2 == 0){
		int y = 1;
		for (int i=0,j=(int)u.size()-1;i<j;i++,j--){
			if (u[i] + u[j] > m) y = 0;
		}
		if (!y){
			if (u.back() <= m) u.pop_back(), U++;
			else Y = 0;
		}
		else U += u.size() / 2;
	}
	if (u.size() % 2){
		int l = 0, r = u.size()-1, a = u.size();
		while (l <= r){
			int mm = (l + r) / 2;
			int y = 1;
			for (int i=0,j=(int)u.size()-1;i<j;i++,j--){
				if (i == mm) i++;
				if (j == mm) j--;
				if (i >= j) break;
				if (u[i] + u[j] > m) y = 0;
			}
			if (u[mm] > m) y = 0;
			if (y){
				a = mm;
				r = mm - 1;
			}
			else l = mm + 1;
		}
		if (a == u.size()) Y = 0;
		else ret = u[a], U += u.size() / 2;
	}
	return ret;
}

int chk(int m)
{
	Y = 1;
	U = 0;
	int l = dfs(1,0,m);
	if (l > m) Y = 0;
	if (l > 0) U++;
	if (U != A) Y = 0;
	return Y;
}

int main()
{
	scanf ("%d",&N);
	for (int i=1;i<N;i++){
		int x,y; scanf ("%d %d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}

	int a = 0;
	for (int i=1;i<=N;i++) a += G[i].size() % 2;
	a /= 2;
	A = a;

	int l = 1, r = N, b = N;
	while (l <= r){
		int m = (l + r) / 2;
		if (chk(m)){
			b = m;
			r = m - 1;
		}
		else l = m + 1;
	}

	printf ("%d %d\n",a,b);

	return 0;
}