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 <stdio.h>
#include <algorithm>
#include <assert.h>
#include <bitset>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <time.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;

#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define ldb ldouble

typedef tuple<int, int, int> t3;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

int IT_MAX = 1 << 19;
int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-10;
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<n;i++)
#define Se second
#define Fi first

const int MX = 200005;

vector<int> T[MX];
int N, m, val = 1;

int dfs(int x, int p = -1){
	vector<int> X;
	for(int c : T[x]){
		if( c == p ) continue;
		X.push_back(dfs(c, x)+1);
	}
	sort(X.begin(), X.end());
	if( !val || X.empty() ) return 0;
	if( X.back() > m ) return val = 0;

	if( X.size()%2 == 0){
		int ch = 1;
		for(int i = 0; i < X.size()/2; i++){
			if( X[i] + X[X.size()-i-1] > m ) ch = 0;
		}
		if( ch ) return 0;
		if( p == -1 ) return val = 0;
		X.pop_back();
	}
	int ch = 1;
	for(int i = 0; i < X.size()/2; i++){
		if( X[i] + X[X.size()-i-2] > m ) ch = 0;
	}
	if( !ch ) val = 0;
	for(int i = X.size()-1; i > 0; i--){
		int op = X.size()/2 < i ? X.size()-i-1 : X.size()-i;
		if( X[i] + X[op] > m ) return X[i];
	}
	return X[0];
}

int main()
{
	int cnt = 1;
	scanf("%d", &N);
	for(int i = 1; i < N; i++){
		int a, b;
		scanf("%d%d", &a, &b);
		T[a].push_back(b);
		T[b].push_back(a);
	}
	for(int i = 1; i <= N; i++){
		if( T[i].size() >= 3 ) cnt += (T[i].size() - 1) / 2;
	}
	int s = 1, e = N;
	while(s <= e){
		m = (s+e) / 2;
		val = 1;
		dfs(1);
		if( val ) e = m-1;
		else s = m+1;
	}swap(s, e);
	printf("%d %d\n", cnt, e);
}