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

typedef long long ll;
const ll MOD = 1e9 + 7;
const int INF = 1 << 30;
vector<int> G[100000];
int dp[100000];

bool check(vector<int>& vec, int m, int B) {
	vector<int> tmp;
	for (int i = 0; i < vec.size(); i++) {
		if (i != m) tmp.push_back(vec[i]);
	}
	int N = tmp.size();
	for (int i = 0; i < N / 2; i++) {
		if (tmp[i] + tmp[N - i - 1] > B) return 0;
	}
	return 1;
}
int solve(int v, int p, int B) {
	if (dp[v] != -1) return dp[v];
	vector<int> child;
	for (int u : G[v]) {
		if (u == p) continue;
		int ret = solve(u, v, B) + 1;
		if (ret > B) {
			return dp[v] = 1 << 30;
		}
		child.push_back(ret);
	}
	if (child.size() == 0) {
		return dp[v] = 0;
	}
	sort(child.begin(), child.end());
	int N = child.size();
	if (N % 2 == 0) {
		bool ok = 1;
		for (int i = 0; i < N / 2; i++) {
			if (child[i] + child[N - i - 1] > B) {
				ok = 0;
				break;
			}
		}
		if (ok) return dp[v] = 0;
		child.pop_back();
		N--;
	}
	int l = -1, r = N - 1;
	while (r - l > 1) {
		int m = (l + r) / 2;
		if (check(child, m, B)) r = m;
		else l = m;
	}
	if (check(child, r, B)) {
		return dp[v] = child[r];
	}
	return dp[v] = 1 << 30;
}
int main() {
	int N;
	cin >> N;
	for (int i = 0; i + 1 < N; i++) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int root = 0;
	int odd = 0;
	for (int i = 0; i < N; i++) {
		if (G[i].size() % 2) {
			odd++;
			root = i;
		}
	}
	int l = -1, r = N;
	while (r - l > 1) {
		int m = (l + r) / 2;
		memset(dp, -1, sizeof(dp));
		solve(root, -1, m);
		if (dp[root] <= m) r = m;
		else l = m;
	}
	cout << odd / 2 << " " << r << endl;
  
  	return 0;
}