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 <bits/stdc++.h>

using namespace std;

int n, cnt, lo, hi, mi, ta, tb;
vector< int > v[100005];
multiset< int > mm;
multiset< int >::iterator it;

int dfs(int a, int b) {
	multiset< int > re = multiset< int >();
	for (int i = 0; i < v[a].size(); i++)
		if (v[a][i] != b) {
			int tt = dfs(v[a][i], a);
			if (tt == -1 || tt + 1 > mi) {
				re.clear();
				return -1;
			}
			re.insert(tt + 1);
		}
	int ret = 0;
	//printf("%d:", a); for (it = re.begin(); it != re.end(); it++) printf(" %d", *it); printf("\n");
	mm = re;
	while (!re.empty()) {
		it = re.end();
		it--;
		ta = *it;
		re.erase(it);
		it = re.upper_bound(mi - ta);
		if (it == re.begin()) {
			if (ret == 0)
				ret = ta;
			else {
				if (b != 0 && mm.size() % 2 == 0) {
					re = mm;
					it = re.end();
					it--;
					re.erase(it);
					ret = 0;
					while (!re.empty()) {
						it = re.end();
						it--;
						ta = *it;
						re.erase(it);
						it = re.upper_bound(mi - ta);
						if (it == re.begin()) {
							if (ret == 0)
								ret = ta;
							else {
								re.clear();
								return -1;
							}
						} else {
							it--;
							re.erase(it);
						}
					}
					//printf("ret %d\n", ret);
					return ret;
				}
				re.clear();
				return -1;
			}
		} else {
			it--;
			re.erase(it);
		}
	}
	//printf("re %d\n", ret);
	return ret;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &ta, &tb);
		v[ta].push_back(tb);
		v[tb].push_back(ta);
	}
	for (int i = 1; i <= n; i++)
		cnt += v[i].size() % 2;
	lo = 1, hi = n + 1;
	while (lo < hi) {
		mi = (lo + hi) / 2;
		//printf("mi %d\n", mi);
		if (dfs(1, 0) != -1)
			hi = mi;
		else
			lo = mi + 1;
	}
	printf("%d %d\n", cnt / 2, lo);
	return 0;
}