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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, D[101000], M;
vector<int>E[101000];
bool Pos(int a, int pp) {
	int i;
	for (i = 0; i < E[a].size(); i++) {
		if (E[a][i] == pp)continue;
		if (!Pos(E[a][i], a))return false;
	}
	vector<int>TP;
	for (i = 0; i < E[a].size(); i++) {
		if (E[a][i] != pp)TP.push_back(D[E[a][i]] + 1);
	}
	sort(TP.begin(), TP.end());
	if (!TP.empty() && TP.back() > M)return false;
	int sz = TP.size();
	if (sz%2 == 0) {
		for (i = 0; i <sz; i++) if (TP[i] + TP[sz - i - 1] > M)break;
		if (i == sz) {
			D[a] = 0;
			return true;
		}
		else {
			if (!pp)return false;
			int b = 0, e = sz - 2, mid, rr = -1;
			while (b <= e) {
				mid = (b + e) >> 1;
				for (i = 0; i < sz - 2; i++) {
					int b = i, e = sz - i - 2;
					if (b >= mid)b++;
					if (e <= mid)e--;
					if (TP[b] + TP[e] > M)break;
				}
				if (i == sz - 2)rr = mid, e = mid - 1;
				else b = mid + 1;
			}
			if (rr == -1)return false;
			D[a] = TP[rr];
			return true;
		}
	}
	else {
		int b = 0, e = sz - 1, mid, rr = -1;
		while (b <= e) {
			mid = (b + e) >> 1;
			for (i = 0; i < sz - 1; i++) {
				int b = i, e = sz - i - 1;
				if (b >= mid)b++;
				if (e <= mid)e--;
				if (TP[b] + TP[e] > M)break;
			}
			if (i == sz - 1)rr = mid, e = mid - 1;
			else b = mid + 1;
		}
		if (rr == -1)return false;
		D[a] = TP[rr];
		return true;
	}
}
int main() {
	int i, a, b, S = 0;
	scanf("%d", &n);
	for (i = 1; i < n; i++) {
		scanf("%d%d", &a, &b);
		E[a].push_back(b);
		E[b].push_back(a);
	}
	for (i = 1; i <= n; i++) {
		S += (E[i].size() - 1) / 2;
	}
	int  e = n, r = 0;
	b = 1;
	while (b <= e) {
		M = (b + e) >> 1;
		if (Pos(1, 0)) {
			r = M;
			e = M - 1;
		}
		else b = M + 1;
	}
	printf("%d %d\n", S + 1, r);
}