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 MAXN = 2e5 + 5;

int n, ansA, leftover[MAXN];
vector<int> adj[MAXN];

int getA(int pos, int prev){
	int sum = 0;
	for(int i : adj[pos]) if(i != prev){
		sum += getA(i, pos);
	}
	if(prev != 0)
		sum += (adj[pos].size()-1)/2;
	else
		sum += (adj[pos].size()+1)/2;
	return sum;
}

bool possible(int pos, int prev, int b){
	vector<int> lens = {};
	// printf("Goto %d\n", pos);
	for(int i : adj[pos]) if(i != prev){
		if(!possible(i, pos, b))
			return false;
		lens.push_back(leftover[i]+1);
	}
	sort(lens.begin(), lens.end());
	// printf("Working with %d (%d)\n", pos, lens.size());
	multiset<int> vacant;
	for(int i = lens.size()-1; i >= 0; --i){
		// printf("(%d)", lens[i]);
		int comp = b-lens[i];
		auto lb = vacant.lower_bound(-comp);
		if(lb != vacant.end()){
			vacant.erase(lb);
		}else{
			vacant.insert(-lens[i]);
		}
	}
	// printf("\n");
	if(vacant.size() > 2)
		return false;
	else if(vacant.size() == 2){
		int u = -(*vacant.begin()); vacant.erase(vacant.begin());
		int v = -(*vacant.begin());
		if(pos != 1 && u <= b)
			leftover[pos] = v; // Case 2 : Cut one leftover, extend the other
		else
			return false;

	}else if(vacant.size() == 1){
		leftover[pos] = -(*vacant.begin());
	}else{
		leftover[pos] = 0;
	}
	return true;
}

bool cekB(int b){
	memset(leftover, 0, sizeof leftover);
	// printf("Ceking %d\n", b);
	return (possible(1, 0, b) && leftover[1] <= b);
}

int main(){
	scanf("%d", &n);
	for(int i = 0; i < n-1; ++i){
		int u, v;
		scanf("%d %d", &u, &v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	ansA = getA(1, 0);

	// TODO: Get B!
	int L = 1, R = n;
	while(L < R){
		int mid = (L+R)/2;
		if(cekB(mid)){
			R = mid;
		}else{
			L = mid+1;
		}
	}
	printf("%d %d\n", ansA, L);
}