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
#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<(int)n;i++)
#define REP(i,n)	for(int i=0;i<=(int)n;i++)
#define srep(i,a,n)	for(int i=a;i<(int)n;i++)
#define SREP(i,a,n)	for(int i=a;i<=(int)n;i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
#define fst	first
#define scn second
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrt(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid*mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return a%b == 0 ? b : gcd(b, a%b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ld eps = 1e-15;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

int n;
vint edge[100010];

int solve(int cur, int parent, int m) {
	vint v;
	rep(i, edge[cur].size()) {
		if (edge[cur][i] == parent)	continue;
		int ret = solve(edge[cur][i], cur, m);
		if (ret == -1)	return -1;
		v.push_back(ret);
	}
	// add self=0(上にも辺があり、そのため次数が奇数）
	if (edge[cur].size() & 1)	v.push_back(0);
	sort(all(v));
	if (parent == -1) {
		for (int i = 0; i < v.size() - i - 1; i++) {
			if (v[i] + v[v.size() - i - 1] > m)	return -1;
		}
		return 0;
	}
	else {
		// 一番大きなv[v.size()-1]を除いてダメならダメ
		for (int i = 0; i < (v.size() - 1) >> 1; i++) {
			if (v[i] + v[v.size() - i - 2] > m)	return -1;
		}
		// 後ろから見て駄目な箇所を探す。それを除くと以降の整合性は取れる。
		for (int i = v.size() - 1; i > v.size() - 1 - i; i--) {
			if (v[i] + v[v.size() - 1 - i] > m)	return v[i] + 1;
		}
		for (int i = (v.size() - 1) >> 1; i > 0; i--) {
			if (v[i] + v[v.size() - i] > m)	return v[i] + 1;
		}
		return v[0] + 1;
	}
}

int main() {
	cin >> n;
	srep(i, 1, n) {
		int a, b;	cin >> a >> b;	a--;	b--;
		edge[a].push_back(b);
		edge[b].push_back(a);
	}
	int A = 0;
	rep(i, n)	A += (edge[i].size() & 1);
	A >>= 1;
	int left = 0, right = n;
	rep(i, 100) {
		int mid = (left + right) >> 1;
		if (solve(0, -1, mid) == -1)	left = mid;
		else	right = mid;
	}
	cout << A << " " << right << endl;
	return 0;
}