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 <vector>
#include <algorithm>
using namespace std;

int n, a[100009], b[100009], sum, G, dp[100009]; vector<int>x[100009]; bool used[100009], flag = false;

int solve(vector<int>F) {
	int res = 0;
	for (int i = 0; i < F.size(); i++) res = max(res, F[i] + F[F.size() - 1 - i]);
	return res;
}

int dfs(int pos) {
	if (flag == true) return (1 << 30);
	used[pos] = true;
	vector<int>I;
	for (int i = 0; i < x[pos].size(); i++) {
		if (used[x[pos][i]] == true) continue;
		I.push_back(dfs(x[pos][i]) + 1);
		if (flag == true) return (1 << 30);
	}
	if (I.size() == 0) {
		dp[pos] = 0;
		return 0;
	}
	sort(I.begin(), I.end());
	if (I[I.size() - 1] > G) {
		flag = true;
		return (1 << 30);
	}
	dp[pos] = (1 << 30);
	if (I.size() % 2 == 0) {
		if (solve(I) <= G) {
			dp[pos] = 0;
			return 0;
		}
	}
	if (x[pos].size() % 2 == 1) I.pop_back();
	if (I.size() % 2 == 0) {
		if (solve(I) <= G) {
			dp[pos] = 0;
			return 0;
		}
	}
	if (I.size() % 2 == 1) {
		int L = 0, R = I.size(), M, maxn = (1 << 30);
		for (int i = 0; i < 20; i++) {
			M = (L + R) / 2;
			vector<int>J; for (int j = 0; j < I.size(); j++) { if (j != M) J.push_back(I[j]); }
			int ret = solve(J);
			if (ret <= G) { R = M; maxn = min(maxn, M); }
			else { L = M; }
		}
		if (maxn != (1 << 30)) {
			dp[pos] = min(dp[pos], I[maxn]);
		}
	}
	if (dp[pos] == (1 << 30)) {
		flag = true;
	}
	return dp[pos];
}

int main() {
	cin >> n;
	for (int i = 1; i <= n - 1; i++) {
		cin >> a[i] >> b[i];
		x[a[i]].push_back(b[i]);
		x[b[i]].push_back(a[i]);
	}
	for (int i = 1; i <= n; i++) { if (x[i].size() % 2 == 1) sum++; }
	int L = 1, R = n + 1, M, minx = (1 << 30);
	for (int i = 0; i < 20; i++) {
		M = (L + R) / 2; G = M;
		for (int j = 1; j <= n; j++) used[j] = false; flag = false;
		int F = dfs(1);
		if (F <= M) { R = M; minx = M; }
		else { L = M; }
	}
	cout << sum / 2 << " " << minx << endl;
	return 0;
}