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 <bits/stdc++.h>
using namespace std;

#define rep(i, N) for (int i = 0; i < N; i++)
#define pb push_back

typedef long long ll;
typedef pair<ll, int> ll_i;
struct edge { int v, w, h; };
const int INF = INT_MAX / 10;
const int MOD = 1e9 + 7;

bool pos(vector<int> a, int ma) {
	int N = a.size();
	bool ok = true;
	rep(i, N / 2) if (a[i] + a[N - 1 - i] > ma) ok = false;
	return ok;
}

int dfs(int u, int p, vector<vector<int>>& G, int ma) {
	vector<int> a;
	for (int v: G[u]) if (v != p) a.pb(dfs(v, u, G, ma));
	sort(a.begin(), a.end());
	int N = a.size();
	if (N % 2) {
		int lb = -1, ub = N;
		while (ub - lb > 1) {
			int mid = (lb + ub) / 2;
			vector<int> b = a;
			b.erase(b.begin() + mid);
			if (pos(b, ma)) ub = mid;
			else lb = mid;
		}
		return ub == N ? INF : a[ub] + 1;
	}
	if (pos(a, ma)) return 1;
	int i;
	for (i = N - 1; i >= 0; i--) if (a[i] <= ma) break;
	if (i == -1) return INF;
	a.erase(a.begin() + i);
	N = a.size();
	if (N % 2) {
		int lb = -1, ub = N;
		while (ub - lb > 1) {
			int mid = (lb + ub) / 2;
			vector<int> b = a;
			b.erase(b.begin() + mid);
			if (pos(b, ma)) ub = mid;
			else lb = mid;
		}
		return ub == N ? INF : a[ub] + 1;
	}
}

int main() {
	int N; cin >> N;
	vector<vector<int>> G(N);
	rep(i, N - 1) {
		int u, v; scanf("%d%d", &u, &v);
		u--, v--;
		G[u].pb(v), G[v].pb(u);
	}
	int A = 1;
	rep(u, N) if (G[u].size() >= 3) A += (G[u].size() - 1) / 2;
	// cout << dfs(0, -1, G, 2) << endl;
	int lb = -1, ub = N;
	int r;
	rep(u, N) if (G[u].size() == 1) r = u;
	while (ub - lb > 1) {
		int mid = (lb + ub) / 2;
		if (dfs(r, -1, G, mid) - 1 <= mid) ub = mid;
		else lb = mid;
	}
	int B = ub;
	cout << A << ' ' << B << endl;
}