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 comment(linker, "/STACK:1000000000")
#include<iostream>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<cassert>

using namespace std;

#define sz(x) (int)(x.size())
#define fi(a,b) for(int i=a;i<b;++i)
#define fj(a,b) for(int j=a;j<b;++j)
#define fo(a,b) for(int o=a;o<b;++o)
#define fdi(a,b) for(int i=a-1;i>=b;--i)
#define fdj(a,b) for(int j=a-1;j>=b;--j)
#define fdo(a,b) for(int o=a-1;o>=b;--o)
#define pb push_back
#define mp make_pair
typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;
////////////////////

int const N = 1e5 + 41;
int const INF = 1e9 + 41;

vector<int> e[N];
int n, d[N], q[N], m;
int a, b;

int getmatch(multiset<int> &t, int x){
	if(sz(t) == 0) return INF;
	multiset<int> :: iterator it = t.upper_bound(x);
	if(it == t.begin()) return INF;
	--it;
	return (*it);
}

void dfs(int x, int p = -1){
	d[x] = 0;
	q[x] = 0;
	vector<int> h;
	fi(0, sz(e[x])){
		int y = e[x][i];
		if(y == p) continue;
		dfs(y, x);
		h.pb(d[y] + 1);
		q[x] += q[y];
	}
	sort(h.begin(), h.end());
	multiset<int> t;
	fdi(sz(h), 0){
		int v0 = h[i];
		int v1 = m - v0;
		int k = getmatch(t, v1);
		if(k + v0 <= m){
			t.erase(t.find(k));
			q[x] += 1;
			continue;
		}
		t.insert(v0);
	}
	q[x] += sz(t);
	if(sz(t) > 0){
		int v0 = (*t.begin());
		if(v0 < m){
			--q[x];
			d[x] = v0;
		}
	}
}

int calc(int m){
	::m = m;
	dfs(0);
	int res = q[0];
	if(d[0] > 0) res += 1;
	return res;
}

void solve(){
	if(n == 2){
		a = 1;
		b = 1;
		return;
	}
	int l = 0;
	int r = N;
	a = calc(n);
	while(r-l>1){
		int m = (l+r)/2;
		if(calc(m) == a) r = m;
		else l = m;
	}
	b = r;
	/*fi(0, n+1){
		if(calc(i) == a){
			b = i;
			break;
		}
	}*/
}

int main(){
#ifdef _DEBUG
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
#endif

	scanf("%d",&n);
	fi(0, n-1){
		int x, y;
		scanf("%d %d",&x,&y);
		--x;--y;
		e[x].pb(y);
		e[y].pb(x);
	}

	solve();

	printf("%d %d\n",a,b);
	return 0;
}