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 <chrono>
#include <random>
#include <cassert>
#include <algorithm>

std::mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());

const int ms = 100100;

std::vector<int> edges[ms];
std::vector<std::pair<int, int>> dp[ms];

void pre(int on, int par) {
	for(int i = 0; i < (int) edges[on].size(); i++) {
		if(edges[on][i] == par) {
			std::swap(edges[on][i], edges[on].back());
			edges[on].pop_back();
			i--;
			continue;
		}
		pre(edges[on][i], on);
	}
	dp[on].resize(edges[on].size());
}

int L;
int solve(const std::vector<std::pair<int, int>> &a, int skip) {
	int ans = 0;
	for(int l = 0, r = (int) a.size() - 1; l < r; l++) {
		if(l == skip) continue;
		if(r == skip) {
			r--;l--;
			continue;
		}
		int s1 = L - a[l].first;
		int s2 = L - a[r].first;
		if(s1 + s2 <= L) {
			r--;
			ans++;
		}
	}
	return ans;
}
std::pair<int, int> dfs(int on) {
	// returns (number of paths, rest of path going up)
	int sum = 0;
	for(int i = 0; i < (int) edges[on].size(); i++) {
		dp[on][i] = dfs(edges[on][i]);
		if(dp[on][i].second == 0) {
			dp[on][i].first++;
			dp[on][i].second = L - 1;
		} else {
			dp[on][i].second--;
		}
		sum += dp[on][i].first;
	}
	// always when possible, pair shit up
	for(auto &p : dp[on]) std::swap(p.first, p.second);
	std::sort(dp[on].begin(), dp[on].end());
	int can = solve(dp[on], 1298739);
	if(2 * can == (int) dp[on].size()) {
		return std::pair<int, int>(sum - can, 0);
	}
	int l = 0, r = (int) dp[on].size() - 1;
	while(l != r) {
		int mid = (l + r + 1) / 2;
		if(solve(dp[on], mid) == can) {
			l = mid;
		} else {
			r = mid - 1;
		}
	}
	assert(solve(dp[on], l) == can);
	return std::pair<int, int>(sum - can, dp[on][l].first);
}

int main() {
	std::ios_base::sync_with_stdio(false); std::cin.tie(NULL);
	int n;
	std::cin >> n;
	for(int i = 1; i < n; i++) {
		int u, v;
		std::cin >> u >> v;
		u--;v--;
		edges[u].push_back(v);
		edges[v].push_back(u);
	}
	pre(0, 0);
	int l = 1, r = n;
	L = n;
	int best = dfs(0).first;
	while(l != r) {
		L = (l + r) / 2;
		if(dfs(0).first == best) {
			r = L;
		} else {
			l = L + 1;
		}
	}
	std::cout << best << ' ' << l << '\n';
}