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 <iostream>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 1e5;
vector<int> g[N];
vector<int> path[N];

void dfs(int u, int p, int x) {
  for (int v : g[u]) if (v != p) {
    dfs(v, u, x);
    path[u].push_back(path[v].back() + 1);
  }

  if (path[u].size() % 2 == 0) {
    path[u].push_back(0);
  }

  sort(path[u].begin(), path[u].end());

  int ok = path[u].size();
  int ng = -1;

  while (ok - ng > 1) {
    int mid = (ok + ng) / 2;

    int mx = 0;
    auto tmp = path[u];
    tmp.erase(tmp.begin() + mid);

    for (int i = 0; i * 2 + 1 < tmp.size(); i++) {
      mx = max(mx, tmp[i] + tmp[tmp.size() - 1 - i]);
    }

    if (mx <= x) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
 
  if (ok == path[u].size()) throw -1;
  swap(path[u][ok], path[u].back());
  
  if (p == -1) {
    if (g[u].size() % 2 == 0 && ok != 0) throw - 1;
    if (g[u].size() % 2 == 1 && path[u].back() > x) throw -1;
  }
}

int main() {
  int n;
  cin >> n;

  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }

  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      ans += (g[i].size() + 1) / 2;
    } else {
      ans += (g[i].size() - 1) / 2;
    }
  }

  int ok = n - 1;
  int ng = -1;
  while (ok - ng > 1) {
    for (int i = 0; i < n; i++) {
      path[i].clear();
    }
    int mid = (ok + ng) / 2;
    try {
      dfs(0, -1, mid);
      ok = mid;
    } catch (...) {
      ng = mid;
    }
  }

  cout << ans << ' ' << ok << endl;
}