#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n, a, b;
  cin >> n;
  vector<vector<int> > adj(n);
  for (auto((i)) = (0); ((i)) < ((n - 1)); ++((i))) {
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int maxC = n - 1;
  set<int> seen;
  vector<int> d(n, -1);
  queue<int> q;
  for (auto((i)) = (0); ((i)) < ((n)); ++((i)))
    if ((adj[i]).size() == 1) {
      if (seen.count(adj[i][0]))
        maxC--;
      else
        seen.insert(adj[i][0]);
      q.push(i);
      d[i] = 0;
    }
  bool allEven = true;
  while (!q.empty()) {
    int curr = q.front();
    q.pop();
    for (int i : adj[curr]) {
      if (d[i] != -1) {
        int len = d[i] + d[curr] + 1;
        if (len > 1 && len & 1) allEven = false;
        continue;
      }
      d[i] = d[curr] + 1;
      q.push(i);
    }
  }
  cout << (allEven ? 1 : 3) << ' ' << maxC << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n, a, b;
  cin >> n;
  vector<vector<int> > adj(n);
  for (auto((i)) = (0); ((i)) < ((n - 1)); ++((i))) {
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int maxC = n - 1;
  set<int> seen;
  vector<int> d(n, -1);
  queue<int> q;
  for (auto((i)) = (0); ((i)) < ((n)); ++((i)))
    if ((adj[i]).size() == 1) {
      if (seen.count(adj[i][0]))
        maxC--;
      else
        seen.insert(adj[i][0]);
      q.push(i);
      d[i] = 0;
    }
  bool allEven = true;
  while (!q.empty()) {
    int curr = q.front();
    q.pop();
    for (int i : adj[curr]) {
      if (d[i] != -1) {
        int len = d[i] + d[curr] + 1;
        if (len > 1 && len & 1) allEven = false;
        continue;
      }
      d[i] = d[curr] + 1;
      q.push(i);
    }
  }
  cout << (allEven ? 1 : 3) << ' ' << maxC << '\n';
}

```