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;
 
#define For(i,a,b) for(int i = a; i <= b; ++ i)
#define pb push_back
#define fi first
#define se second
 
const int N = 1e5 + 5;
typedef pair < int, int > ii;

int t, n, A = 1, newA;
vector < int > adj[N];
 

int dfs(int u, int pa, int lim) {
    multiset < int > s;
    multiset < int > :: iterator it;
    for (int i = 0; i < adj[u].size() ; ++ i){
		int v = adj[u][i];
		if (v == pa) continue;
        int cur = dfs(v, u, lim) + 1; s.insert(cur);
    }
    vector < int > data;
    while(s.size()) {
        it = s.end(); --it; int x = (*it);
        s.erase(it);
        it = s.upper_bound(lim - x);
        if (it == s.begin()) data.pb(x);
        else --it, s.erase(it), ++newA;
    }
    reverse(data.begin(), data.end());
    while(data.size() && data.back() == lim) ++ newA, data.pop_back();
    if (data.empty()) return 0;
    else { newA += data.size() - 1; return data[0]; }
}
bool check(int lim){
	newA = 0;
    newA += dfs(1, 1, lim) != 0;
    return newA <= A;
}
signed main(){
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n;
	For(i, 1, n - 1){
		int u, v;
		cin >> u >> v;
		adj[u].pb(v);  adj[v].pb(u);
	}
	For(i, 1, n)  A += (adj[i].size() - 1) / 2;
	int l = 1, r = n, ans;
	while(l <= r){
		int mid = (l + r)/2;
		if(check(mid)) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	cout << A << ' ' << ans;
}