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 <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;

int n;
vector<int> gph[MAXN];
vector<int> dfn;

int dp[MAXN];

int query(int k){
	for(auto &i : dfn){
		if(i == 1) break;
		vector<int> v;
		for(auto &j : gph[i]) v.push_back(dp[j] + 1);
		sort(v.begin(), v.end());
		auto trial_pos = [&](int x){
			int l = 0, r = v.size() - 1;
			for(int i=0; i<v.size()/2; i++){
				if(l == x) l++;
				if(r == x) r--;
				if(v[l] + v[r] > k) return 0;
				l++;
				r--;
			}
			return 1;
		};
		if(v.size() % 2 == 0){
			v.insert(v.begin(), 0);
		}
		int s = 0, e = v.size();
		while(s != e){
			int m = (s+e)/2;
			if(trial_pos(m)) e = m;
			else s = m + 1;
		}
		if(s == v.size() || v[s] >= k) return 0;
		dp[i] = v[s];
	}
	vector<int> v;
	for(auto &i : gph[1]) v.push_back(dp[i] + 1);
	sort(v.begin(), v.end());
	if(v.back() > k) return 0;
	if(v.size() % 2 == 1) v.pop_back();
	for(int i=0; i<v.size()/2; i++){
		if(v[i] + v[v.size()-i-1] > k) return 0;
	}
	return 1;
}

void dfs(int x){
	dfn.push_back(x);
	for(auto &i : gph[x]){
		gph[i].erase(find(gph[i].begin(), gph[i].end(), x));
		dfs(i);
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1; i<n; i++){
		int s, e;
		scanf("%d %d",&s,&e);
		gph[s].push_back(e);
		gph[e].push_back(s);
	}
	dfs(1);
	int ans = n - 1;
	for(int i=1; i<=n; i++){
		ans -= ((i != 1) + gph[i].size()) / 2;
	}
	int s = 1, e = n;
	reverse(dfn.begin(), dfn.end());
	while(s != e){
		int m = (s+e)/2;
		if(!query(m)) s = m+1;
		else e = m;
	}
	printf("%d %d\n", ans, s);
}


