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 <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////
ll N;
vvll g;

ll dfs(ll u, ll v, ll x) {
	vll a;
	for (ll w : g[v]) {
		if (u == w) { continue; }
		ll k = dfs(v, w, x);
		//cout << w << " " << x << " " << k << endl;

		if (k == -1) { return -1; }
		a.push_back(k + 1);
	}
	sort(a.begin(), a.end());
	ll M = a.size();
	rep(i, M) {
		if (a[i] > x) { return -1; }
	}

	if (g[v].size() % 2 == 1) {
		bool b = 1;
		rep(i, M) { if (a[i] + a[M - 1 - i] > x) { b = 0; } }
		if(b)return 0;

		a.pop_back();
		M--;
	}
	

	rep(i, M - 1) { if (a[i] + a[M - 2 - i] > x) { return -1; } }
	
	rep(i, M / 2) {
		if (a[i] + a[M - 1 - i] > x) {return a[M - 1 - i];}
	}
	rep(i, M / 2) {
		ll j = M / 2 - i;
		if (a[j] + a[M - j] > x) { return a[j]; }
	}
	return a[0];
}

bool ch(ll x) {
	ll leaf = 0;
	repn(i, N) { if (g[i].size() == 1) { leaf = i; break; } }
	ll t = dfs(leaf, g[leaf][0], x);
	if (t!=-1&&t + 1 <= x) { return true; }
	else { return false; }
}
int main() {
	cin >> N;
	g.resize(N + 1);
	repn(i, N-1) {
		ll a, b;
		cin >> a >> b;
		g[a].push_back(b);
		g[b].push_back(a);
	}

	ll cnt = 0;
	repn(i, N) { if (g[i].size() % 2 == 1) { cnt++; } }
	cout << cnt / 2 <<" ";

	ll l = 0;
	ll r = N;

	while (r - l > 1) {
		ll m = (r + l) / 2;
		if (ch(m)) { r = m; }
		else { l = m; }
	}

	cout << r << endl;
}