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;
vector<vector<int> > gr;
bool dfs(int v, int p, int h) {
  if (gr[v].size() == 1 && h % 2) {
    return false;
  }
  for (int u : gr[v]) {
    if (u == p) continue;
    if (!dfs(u, v, h + 1)) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  gr.resize(n);
  for (int i = 1; i < n; ++i) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    gr[v].push_back(u);
    gr[u].push_back(v);
  }
  int max_diff = n - 1;
  for (int i = 0; i < n; ++i) {
    int cnt = 0;
    for (int v : gr[i]) {
      if (gr[v].size() == 1) ++cnt;
    }
    if (cnt) {
      max_diff -= cnt - 1;
    }
  }
  int min_diff = 0;
  for (int i = 0; i < n; i++) {
    if (gr[i].size() > 1) continue;
    min_diff = (dfs(i, -1, 0) ? 1 : 3);
    break;
  }
  cout << min_diff << " " << max_diff << "\n";
}
