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
//#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
//using namespace atcoder;
#define int long long
   template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
    const int MAX =1002;
const int MOD =1000000007;
vector<int> hen[100000];
bool used[100000];
int ok,ng,mid;
int dfs(int a,bool a0){
  used[a]=true;
  vector<int> v;
  for(auto&& i:hen[a])if(!used[i]){
    int b=dfs(i,false);
    if(b==-1)return -1;
    v.push_back(b+1);
  }
  sort(v.begin(),v.end());
  if(v.size()%2==0){
    bool t=true;
      for(int i=0;i*2<v.size();i++){
        if(v[i]+v[v.size()-1-i]>mid){
          t=false;
          break;
        }
      }
    if(t)return 0;
    if(v[v.size()-1]>mid||a0)return -1;
    v.pop_back();
  }
  
    int ok2=v.size(),ng2=-1,mid2;
    while(abs(ok2-ng2)>1){
      mid2=(ok2+ng2)/2;
      bool t=true;
      for(int i=0;i*2<v.size()-1;i++){
        if(v[i+(i>=mid2)]+v[v.size()-2-i+(v.size()-2-i>=mid2)]>mid){
          t=false;
          break;
        }
      }
      if(t)ok2=mid2;
      else ng2=mid2;
    }
    if(ok2==v.size())return -1;
    return v[ok2];
    
  
}
signed main(){
 int n,a,b,ansa=1;
 cin>>n;
 for(int i=0;i<n-1;i++){
   cin>>a>>b;
   a--;b--;
   hen[a].push_back(b);
   hen[b].push_back(a);
 }
 for(int i=0;i<n;i++)ansa+=(hen[i].size()-1)/2;
  ng=0;ok=n;
  while(abs(ok-ng)>1){
    mid=(ok+ng)/2;
    fill(used,used+n,false);
    int y=dfs(0,true);
    if(y==-1||y>mid)ng=mid;
    else ok=mid;
  }
  cout<<ansa<<' '<<ok<<endl;
 
}