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 {{{
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <complex>
#include <algorithm>
using namespace std;
// }}}
// #define {{{
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
// }}}

const int N = 1e5 + 10;
int n;
vi g[N];

bool check(vi&f,int B,int skip) {
  for(int i=0,j=sz(f)-1;i<j;++i,--j) {
    if(i == skip) ++i;
    if(j == skip) --j;
    if(i < j && f[i] + f[j] > B)
      return false;
  }
  return true;
}

int dfs(int c,int fa,int B) {
  vi f;
  for(auto t : g[c]) if(t != fa) {
    f.pb(dfs(t , c , B));
  }
  if(~sz(f) & 1) f.pb(0);
  sort(all(f));
  if(!check(f , B , sz(f) - 1))
    throw 1;
  int l = -1 , r = sz(f) - 1;
  while(l + 1 < r) {
    int mid = (l + r) >> 1;
    (check(f , B , mid) ? r : l) = mid;
  }
  return f[r] + 1;
}

bool check(int B,int rt) {
  try {
    int d = dfs(rt , 0 , B);
    if(d > B + 1)
      return false;
    return true;
  } catch(...) {
    return false;
  }
}

int main(){
  scanf("%d",&n);
  rep(i,1,n) {
    int x,y;
    scanf("%d%d",&x,&y);
    g[x].pb(y);
    g[y].pb(x);
  }
  int A = 0;
  rep(i,1,n+1) A += sz(g[i]) & 1;
  int rt = 1;
  while(~sz(g[rt]) & 1) rt++;
  A >>= 1;
  int l = 0 , r = n;
  while(l + 1 < r) {
    int mid = (l + r) >> 1;
    (check(mid , rt) ? r : l) = mid;
  }
  printf("%d %d\n",A,r);
  return 0;
}