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;

int N;
vector< int > g[100000];
int dp[100000];

bool rec(int idx, int par, int length)
{
  vector< int > vs;
  for(auto &to : g[idx]) {
    if(to == par) continue;
    if(!rec(to, idx, length)) return (false);
    if(dp[to] + 1 > length) return (false);
    vs.push_back(dp[to] + 1);
  }
  sort(begin(vs), end(vs));

  auto check = [&](int idx)
  {
    deque< int > st;
    for(int i = 0; i < vs.size(); i++) {
      if(i == idx) continue;
      st.push_back(vs[i]);
    }
    while(!st.empty()) {
      if(st.back() + st.front() > length) return (false);
      st.pop_back();
      st.pop_front();
    }
    return (true);
  };

  if(vs.size() % 2 == 0) {
    if(check(-1)) {
      dp[idx] = 0;
      return (true);
    }
    vs.pop_back();
  }

  int ok = (int) vs.size(), ng = -1;
  while(ok - ng > 1) {
    int mid = (ok + ng) / 2;
    if(check(mid)) ok = mid;
    else ng = mid;
  }
  if(ok == vs.size()) return (false);
  dp[idx] = vs[ok];
  return (true);
}

bool check(int v)
{
  int root = -1;
  for(int i = 0; i < N; i++) {
    if(g[i].size() & 1) root = i;
  }
  return (rec(root, -1, v));
}

int main()
{
  cin >> N;
  for(int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }

  int A = 1;
  for(int i = 0; i < N; i++) {
    A += g[i].size() & 1;
  }

  int ok = N, ng = 0;
  while(ok - ng > 1) {
    int mid = (ok + ng) / 2;
    if(check(mid)) ok = mid;
    else ng = mid;
  }

  cout << A / 2 << " " << ok << endl;
}