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:
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx,avx2,popcnt")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
using namespace std;
using ll = long long;
using ld = long double;
const int maxn = 1<<17, mod = 1e9 + 7;
int n, _x;
vector<int> g[maxn];
int B = 0, A = 0;
int get(vector<int> &a, int s) {
	int l = 0, r = a.size()-1, mx = 0;
	if(l == s) l++;
	if(r == s) r--;
	while(l < r) {
		mx = max(mx, a[l] + a[r]);
		l++, r--;
		if(l == s) l++;
		if(r == s) r--;
	}
	return mx;
}
int dfs(int v, int p, int h = 0) {
	vector<int> ch;
	if(g[v].size()&1) ch.pb(h);
	for(auto i : g[v]) if(i != p) {
		int t = dfs(i, v, h+1);
		if(t == -1) return -1;
		if(t) ch.pb(t);
	}
	sort(all(ch));
	if(ch.size()%2 == 0) {
		return get(ch, -1)-2*h <= _x ? 0 : -1;
	}
	p = 0;
	for(int i = 1<<17; i>>=1;)
		if(p+i < ch.size() && get(ch, p+i)-2*h > _x) p += i;
	p += (get(ch, p)-2*h > _x);
	return p >= ch.size() ? -1 : ch[p];
}
bool can(int x) {
	_x = x;
	return dfs(1, 1) != -1;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n;
	for(int f, t, i = 1; i < n; i++) {
		cin >> f >> t;
		g[f].pb(t);
		g[t].pb(f);
	}
	for(int i = 1; i <= n; i++) A += g[i].size()&1; A/=2;
	for(int i = 1<<17; i>>=1;) if(!can(B+i)) B += i;
	cout << A << " " << B+1 << '\n';
	return 0;
}