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 <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

vector<int> g[100000];
int c[100000];
bool used[100000];
bool nuee;
int d;
void dfs(int x){
  if(nuee) return;
  used[x]=1;
  vector<int> v;
  for(auto y:g[x]){
    if(used[y]) continue;
    dfs(y);
    if(nuee) return;
    v.push_back(c[y]+1);
  }
  if(v.empty()) return;
  if(v.size()==1){
    if(v[0]>d) nuee=1;
    c[x]=v[0];
    return;
  }
  if(v.size()%2==0) v.push_back(0);
  sort(v.begin(), v.end());
  int m=v.size();
  bool ok0=0;
  for(int i=0; i<m; i++){
    int c1=0, c2=m-1;
    bool ok=1;
    for(int j=0; j<m/2; j++){
      if(c1==i) c1++;
      if(c2==i) c2--;
      if(v[c1]+v[c2]>d){
        ok=0;
        break;
      }
      c1++; c2--;
    }
    if(ok){
      ok0=1;
      c[x]=v[i];
      break;
    }
  }
  if(!ok0){
    nuee=1;
  }
}

int main()
{
	int n; cin>>n;
  for(int i=0; i<n-1; i++){
    int a, b;
    cin>>a>>b;
    a--; b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int ct=0;
  int v0;
  for(int i=0; i<n; i++){
    if(g[i].size()%2) ct++;
    if(g[i].size()==1) v0=i;
  }
  cout<<ct/2<<" ";
  int d1=1, d2=2e5;
  while(d1!=d2){
    d=(d1+d2)/2;
    fill(c, c+n, 0);
    fill(used, used+n, 0);
    nuee=0;
    dfs(v0);
    if(nuee) d1=d+1;
    else d2=d;
  }
	cout<<d1<<endl;
	return 0;
}