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
#ifndef LOCAL
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const int MAXN = 120000;
int gfl = 0;

vector<int> eds[MAXN];
int n;

int dfs1(int v, int x, int p) {
	vector<int> vv;
	for (int u: eds[v]) {
		if (u == p)
			continue;
		vv.push_back(dfs1(u, x, v) + 1);
	}
	sort(vv.begin(), vv.end());
	for (int u: vv)
		if (u > x) {
			gfl = 1;
			return 0;
		}
	if (vv.size() % 2 == 0) {
		int fl = 0;
		for (int i = 0; i < vv.size(); ++i)
			if (vv[i] + vv[vv.size() - 1 - i] > x) {
				fl = 1;
				break;
			}
		if (!fl)
			return 0;
	}
	if ((p == -1 && vv.size() % 2 == 1) || (p != -1 && vv.size() % 2 == 0)) {
		if (!vv.empty())
			vv.pop_back();
		else
			vv.push_back(0);
	}
	if (p == -1) {
		for (int i = 0; i < vv.size(); ++i) {
			if (vv[i] + vv[vv.size() - 1 - i] > x) {
				gfl = 1;
				return 0;
			}
		}
		return 0;
	}
	else {
		for (int i = 0; i < (int)vv.size() - 1; ++i) {
			if (vv[i] + vv[vv.size() - 2 - i] > x) {
				gfl = 1;
				return 0;
			}
		}
		int cur = vv.size() - 1;
		while (cur > 0) {
			int oth = vv.size() - 2 - (cur - 1);
			if (oth >= cur)
				++oth;
			if (vv[cur] + vv[oth] > x)
				break;
			--cur;
		}
		return vv[cur];
	}
}

int main() {
	cin >> n;
	for (int i = 0; i < n - 1; ++i) {
		int a, b;
		cin >> a >> b;
		--a, --b;
		eds[a].push_back(b);
		eds[b].push_back(a);
	}
	int l = 0;
	int r = n;
	while (r - l > 1) {
		int mid = (l + r) >> 1;
		gfl = 0;
		dfs1(0, mid, -1);
		if (gfl)
			l = mid;
		else
			r = mid;
	}
	int a = 0;
	for (int i = 1; i < n; ++i) {
		int c = eds[i].size() - 1;
		if (c)
			--c;
		a += (c + 1) / 2;
	}
	int c = eds[0].size();
	a += (c + 1) / 2;
	cout << a << " " << r << "\n";
	return 0;
}


