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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
const int oo = 1e9;
const int MOD = 1000000007;
const int N = 300010;
#define x first
#define y second
#define pb push_back
#define mp make_pair

int n, u, v, lng, num, md;
vector<int> g[N];
pii fn;

int dfs(int u, int p) {
	multiset<int> st;
	int nt = 0;
	for (int i = 0; i < g[u].size(); ++i) {
		int v = g[u][i];
		if (v == p)continue;
		++nt;
		st.insert(1 + dfs(v, u));
	}
	num += nt/2;
	vector<int> go;
	while (st.size()) {
		int bk = (*--st.end());
		st.erase(st.find(bk));
		auto it = st.upper_bound(md - bk);
		if (it != st.begin())
			it--;
		if (st.size() && *it + bk <= md) {
			lng = max(lng, *it + bk);
			st.erase(it);
		} else {
			go.pb(bk);
			lng = max(lng, bk);
		}
	}
	sort(go.begin(),go.end());
	if(p == -1 && nt%2 == 0 && go.size())lng = oo;
	if(nt%2 == 0 && go.size() > 2)lng = oo;
	if(nt%2 == 1 && go.size() > 1)lng = oo;
	if(!go.size())return 0;
	return go[0];
}

bool check() {
	lng = num = 0;
	int res = dfs(1, -1);
	lng = max(lng, res);
	num += (res>0);
	return lng <= md;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n - 1; ++i) {
		scanf("%d%d", &u, &v);
		g[u].pb(v);
		g[v].pb(u);
	}
	int lo = 1 , hi = 300000, best = -1;
	while (lo <= hi) {
		md = (lo + hi) / 2;
		bool ss = check();
		if (ss) {
			fn = mp(num, lng);
			hi = md - 1;
		} else {
			lo = md + 1;
		}
	}
	printf("%d %d\n", fn.x, fn.y);
	return 0;
}