Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

7
1 2
2 3
2 4
4 5
4 6
6 7

Sample Output 1

3 2

We can make a Christmas Tree as shown in the figure below:

Sample Input 2

8
1 2
2 3
3 4
4 5
5 6
5 7
5 8

Sample Output 2

2 5

Sample Input 3

10
1 2
2 3
3 4
2 5
6 5
6 7
7 8
5 9
10 5

Sample Output 3

3 4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#ifdef SG
	#include <debug.h>
#else
	template<typename T> struct outputer;
	struct outputable {};
	#define PRINT(...)
	#define OUTPUT(...)
	#define show(...)
	#define debug(...)
	#define deepen(...)
	#define timer(...)
	#define fbegin(...)
	#define fend
	#define pbegin(...)
	#define pend
#endif

#define ARG4(_1,_2,_3,_4,...) _4

#define forn3(i,l,r) for (int i = int(l); i < int(r); ++i)
#define forn2(i,n) forn3 (i, 0, n)
#define forn(...) ARG4(__VA_ARGS__, forn3, forn2) (__VA_ARGS__)

#define ford3(i,l,r) for (int i = int(r) - 1; i >= int(l); --i)
#define ford2(i,n) ford3 (i, 0, n)
#define ford(...) ARG4(__VA_ARGS__, ford3, ford2) (__VA_ARGS__)

#define ve vector
#define pa pair
#define tu tuple
#define mp make_pair
#define mt make_tuple
#define pb emplace_back
#define fs first
#define sc second
#define all(a) (a).begin(), (a).end()
#define sz(a) ((int)(a).size())

typedef long double ld;
typedef int64_t ll;
typedef uint64_t ull;
typedef uint32_t ui;
typedef uint16_t us;
typedef uint8_t uc;
typedef pa<int, int> pii;
typedef pa<int, ll> pil;
typedef pa<ll, int> pli;
typedef pa<ll, ll> pll;
typedef ve<int> vi;

template<typename T> inline auto sqr (T x) -> decltype(x * x) {return x * x;}
template<typename T1, typename T2> inline bool umx (T1& a, T2 b) {if (a < b) {a = b; return 1;} return 0;}
template<typename T1, typename T2> inline bool umn (T1& a, T2 b) {if (b < a) {a = b; return 1;} return 0;}

const int N = 100000;

struct Input {
	int n;
	vi e[N];
	
	bool read () {
		if (!(cin >> n)) {
			return 0;
		}
		forn (i, n - 1) {
			int x, y;
			scanf("%d%d", &x, &y);
			--x;
			--y;
			e[x].pb(y);
			e[y].pb(x);
		}
		return 1;
	}

	void init (const Input &input) {
		*this = input;
	}
};

struct Data: Input {
	int ans_a, ans_b;
	
	void write () {
		cout << ans_a << ' ' << ans_b << endl;
	}
};


namespace Main {
	
	struct Solution: Data {
		int root;
		int mxl;
		
		int rec (int v, int pr) {
			vi q;
			for (int to : e[v]) {
				if (to == pr) {
					continue;
				}
				int val = rec(to, v);
				if (val == -1 || val + 1 > mxl) {
					return -1;
				}
				q.pb(val + 1);
			}
			sort(all(q), greater<int>());
			
			if (!sz(q)) {
				return 0;
			}
			
			int val0 = q[0];
			if (sz(q) % 2 == 0) {
				q[0] = -1;
			}
			
			int j = 0;
			forn (i, sz(q)) {
				if (q[i] == -1) {
					continue;
				}
				while (j < sz(q) && (j <= i || q[j] == -1 || q[i] + q[j] > mxl)) {
					++j;
				}
				if (j < sz(q)) {
					assert(q[i] != -1);
					assert(q[j] != -1);
					q[i] = q[j] = -1;
				}
			}
			
			vi qq;
			if (sz(q) % 2 == 0) {
				qq.pb(val0);
			}
			forn (i, sz(q)) {
				if (q[i] != -1) {
					qq.pb(q[i]);
				}
			}
			if (sz(qq) == 2 && qq[0] + qq[1] <= mxl) {
				qq.clear();
			}
			
			if (sz(qq) == 0) {
				return 0;
			} else if (sz(qq) == 1) {
				return qq[0];
			} else if (sz(qq) == 2) {
				return qq[1];
			} else {
				return -1;
			}
		}
	
		bool check (int val) {
			mxl = val;
			return rec(root, -1) != -1;
		}
		
		void solve () {
			root = 0;
			while (sz(e[root]) > 1) {
				root++;
			}
			
			ans_a = 0;
			forn (i, n) {
				if (sz(e[i]) & 1) {
					ans_a++;
				}
			}
			ans_a /= 2;
			
			umx(ans_a, 1);
			
			int lb = 0, rb = 1000000000;
			while (lb < rb) {
				int md = (lb + rb) / 2;
				if (check(md)) {
					rb = md;
				} else {
					lb = md + 1;
				}
			}
			ans_b = lb;
		}
		
		void clear () {
			*this = Solution();
		}
	};
}


Main::Solution sol;

int main () {
	cout.setf(ios::showpoint | ios::fixed);
	cout.precision(20);

	#ifdef SG
		freopen((problemname + ".in").c_str(), "r", stdin);
//		freopen((problemname + ".out").c_str(), "w", stdout);
		while (sol.read()) {
			sol.solve();
			sol.write();
			sol.clear();
		}
	#else
		sol.read();
		sol.solve();
		sol.write();
	#endif
	
	/*
	int t;
	cin >> t;
	forn (i, t) {
		sol.read();
		sol.solve();
		sol.write();
		sol.clear();
	}
	*/
	
	return 0;
}