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 pb push_back
#define mp make_pair
#define st first
#define nd second
typedef pair < int , int > pp;
const int mod = 1e9 + 7;
const int N   = 1e5 + 5;

vector < int > V[N];
int T[N],t,ok,K;

int f(int x, int p){
    if(ok == 0) return 0;
    vector < int > v;
    if(V[x].size() % 2 == 1) v.pb(0);
    for(int y, i=0; i<V[x].size(); i++)
        if((y = V[x][i]) != p){
            v.pb( f(y,x)+1 );
            if(ok == 0) return 0;
        }
    sort(v.begin() , v.end());
    if(x == 1){
        t = v.size();
        for(int i=0; i<t; i++)
            if(v[i] + v[t-i-1] > K) return ok=0;
        return 0;
    }
    int i,l,m,r;
    l = 0;
    r = v.size();
    for(;l < r;){
        m = l+r >> 1;
        t = 0;
        for(i=0;i<v.size();i++)
            if(i != m)
                T[++t] = v[i];
        
        for(i=1;i<=t/2;i++)
            if(T[i] + T[t-i+1] > K)
                break;
        if(i <= t/2) l = m+1;
        else r = m;
    }
    if(l == v.size()) return ok=0;
    return v[l];
}

int n,i,x,y,a,l,m,r;
int main(){
    cin >> n;
    for(i=1;i<n;i++){
        scanf("%d%d",&x,&y);
        V[x].pb(y); V[y].pb(x);
    }
    
    a = 1;
    for(i=1;i<=n;i++)
        a += (int)V[i].size()-1 >> 1;
        
    l = n/a-1;
    r = n;
    for(; l<r ;){
        m = l+r >> 1;
        ok = 1;
        K = m;
        f(1,-1);
        if(ok) r = m;
        else l = m+1;
    }
    
    cout << a << " " << l << endl;
    return 0;
}