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;

const int N = 100005;

int n, cnt;
int f[N];
int sum[N];
bool flag;
vector<int> G[N];

void dfs(int u, int p, int x) {
	vector<int> go;
	for (auto v : G[u]) {
		if (v == p) continue;
		dfs(v, u, x), go.push_back(f[v] + 1);
	}
	int sz = go.size();
	bool even = 0;
	cnt += sz / 2;
	if (sz % 2 == 0) go.push_back(0), sz++, even = 1;
	else if (u == 1) cnt++;
	sort(go.begin(), go.end());
	// bubbletea prevoi 2017
	for (int i = 0; i < sz; ++i) sum[i] = 0;
	for (int i = 0; i <= sz / 2; ++i) {
        if (i < sz - i - 2 && sz - i - 2 < sz) {
            if (go[i] + go[sz - i - 2] > x) sum[sz - i - 1]++;
        }
        if (i < sz - i - 1 && sz - i - 1 < sz) {
            if (go[i] + go[sz - i - 1] > x) sum[i + 1]++, sum[sz - i - 1]--;
        }
        if (i < sz - i && sz - i < sz) {
            if (go[i] + go[sz - i] > x) sum[0]++, sum[i]--;
        }
	}
	for (int i = 1; i < sz; ++i) sum[i] += sum[i - 1];
	for (int i = 0; i < sz; ++i) {
		if (u == 1 && even && go[i]) continue;
		if (!sum[i] && go[i] <= x) { f[u] = go[i]; return; }
	}
	flag = 1;
}

int check(int x) {
	flag = cnt = 0, dfs(1, 1, x); 
	return (!flag) ? cnt : 0;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int u, v; cin >> u >> v;
		G[u].push_back(v), G[v].push_back(u);
	}
	int l = 1, r = n;
	while (l < r) {
		int mid = (l + r) >> 1;
		if (check(mid)) r = mid; else l = mid + 1;
	}
	cout << check(l) << ' ' << l;
}