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 main() {
   ios_base::sync_with_stdio(false);
   int N;
   cin >> N;
   vector<vector<int>> adj(N);
   for (int i = 0; i < N - 1; ++i) {
      int v, u;
      cin >> v >> u;
      --v, --u;
      adj[v].emplace_back(u);
      adj[u].emplace_back(v);
   }
   int ans = 0;
   for (int i = 0; i < N; ++i) {
      ans += int(adj[i].size()) & 1;
   }
   ans >>= 1;
   int low = 1, high = N;
   while (low < high) {
      int mid = (low + high) >> 1;
      int cans = 0;
      function<int(int, int)> dfs = [&](int v, int p) {
         multiset<int> nxts;
         for (int u : adj[v]) {
            if (u != p) {
               nxts.emplace(dfs(u, v) + 1);
            }
         }
         vector<int> bigs;
         while (nxts.size()) {
            int x = *(--nxts.end());
            nxts.erase(nxts.find(x));
            auto it = nxts.upper_bound(mid - x);
            if (it == nxts.begin()) {
               bigs.emplace_back(x);
            } else {
               nxts.erase(--it);
               cans++;
            }
         }
         reverse(bigs.begin(), bigs.end());
         while (bigs.size() && bigs.back() == mid) {
            cans++;
            bigs.pop_back();
         }
         if (bigs.empty()) return 0;
         cans += int(bigs.size()) - (v > 0);
         return bigs[0];
      };
      dfs(0, -1);
      if (ans == cans) {
         high = mid;
      } else {
         low = mid + 1;
      }
   }
   cout << ans << " " << low << "\n";
   return 0;
}