You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:

  * For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0. 



Note that you can put very large positive integers (like 10^{(10^{10})}).

It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.

<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).

<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0. 

What are the minimum and the maximum possible values of f for the given tree? Find and print both.

Input

The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.

The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.

Output

Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.

Examples

Input


6
1 3
2 3
3 4
4 5
5 6


Output


1 4


Input


6
1 3
2 3
3 4
4 5
4 6


Output


3 3


Input


7
1 2
2 7
3 4
4 7
5 6
6 7


Output


1 6

Note

In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3. 

<image>

In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> tree(n);
  for (int i = 0; i + 1 < n; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  bool odd = false;
  int total = 0;
  int r = 0;
  function<int(int, int, int)> dfs = [&](int s, int p, int par) {
    if (s != r && tree[s].size() == 1) {
      if (par) odd = true;
      return 0;
    }
    int ans = 1000000000;
    for (auto u : tree[s]) {
      if (u == p) {
        continue;
      }
      int cur = dfs(u, s, par ^ 1) + 1;
      if (cur < ans) swap(ans, cur);
      if (cur > 1 && cur < 1000000000) {
        total += cur;
      }
    }
    return ans;
  };
  while (tree[r].size() > 1) ++r;
  auto res = dfs(r, -1, 0);
  if (res == 2)
    total++;
  else
    total += res;
  cout << (odd ? 3 : 1) << " " << total << '\n';
  return 0;
}
