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 N = 100005;

int n, lim, ans, dt[N];

vector<int> adj[N], arr;

bool can (int C, int P) {
	for(auto &T : adj[C]) {
		if(T == P) continue;
		if(!can(T, C)) return false;
	}
	arr.clear();
	for(auto &T : adj[C]) {
		if(T == P) continue;
		if(dt[T]+1 > lim) return false;
		arr.push_back(dt[T]+1);
	}
	sort(arr.begin(), arr.end());
	int N = arr.size();
	if(N % 2 == 0) {
		bool F = false;
		for(int i=0;i<N/2;i++) {
			if(arr[i] + arr[N-1-i] > lim) F = true;
		}
		if(!F) {dt[C] = 0; return true;}
		arr.pop_back(); N--;
	}
	int S = 0, E = N;
	while(S<E) {
		int M = (S+E)/2, L = 0, R = N-1;
		bool F = false;
		while(1) {
			if(L == M) L++;
			if(R == M) R--;
			if(L >= R) break;
			if(arr[L] + arr[R] > lim) {F = true; break;}
			L++; R--;
		}
		F ? S = M+1 : E = M;
	}
	if(S == N) return false;
	dt[C] = arr[S];
	return true;
}

bool can () {
	if(!can(1, 0)) return false;
	return (adj[1].size() % 2 || dt[1] == 0);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++) {
		int A, B;
		scanf("%d%d",&A,&B);
		adj[A].push_back(B);
		adj[B].push_back(A);
	}
	ans = n-1;
	for(int i=1;i<=n;i++) {
		ans -= (int)adj[i].size() / 2;
	}
	printf("%d ",ans);
	int S = 0, E = n;
	while(S<E) {
		lim = (S+E)/2;
		can() ? E = lim : S = lim+1;
	}
	printf("%d\n",S);
}