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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcountll

#define INF 1e16
#define mod 1000000007

int N;
vector<int> g[100010];
int d[100010];
int dp[100010];

bool dfs(int v,int pre,int X){
  vector<int> w;
  for(int nv : g[v]){
    if(nv==pre)continue;
    if(!dfs(nv,v,X))return false;
    w.push_back(dp[nv]+1);
  }
  if(d[v])w.push_back(0);
  sort(all(w));
  int m=w.size();
  if(m%2==0){ // r
    rep(j,m)if(w[j]+w[m-1-j]>X)return false;
  }else{
    int lb=-1,ub=m;
    while(ub-lb>1){
      int mid=(lb+ub)/2;
      bool ok=true;
      int l=0,r=m-1;
      if(l==mid)l++;
      if(r==mid)r--;
      while(l<r){
        if(w[l]+w[r]>X)ok=false;
        l++; if(l==mid)l++;
        r--; if(r==mid)r--;
      }
      if(ok) ub=mid;
      else lb=mid;
    }
    if(ub==m)return false;
    dp[v]=w[ub];
  }
  return true;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  cin>>N;
  rep(i,N-1){
    int a,b;
    cin>>a>>b;
    a--;b--;
    g[a].push_back(b);g[b].push_back(a);
  }
  memset(dp,-1,sizeof(dp));

  int A=0;
  rep(i,N){
    if((int)g[i].size()%2==1)A++;
    d[i]=(int)g[i].size()%2;
  }
  A/=2;

  int lb=0,ub=N;
  while(ub-lb>1){
    int mid=(lb+ub)/2;
    if(dfs(0,-1,mid))ub=mid;
    else lb=mid;
  }
  cout<<A<<" "<<ub<<endl;

  return 0;
}