#include <bits/stdc++.h>
using namespace std;
int dfs(vector<vector<int>>& adj, vector<bool>& vis, int node) {
  int result = -2;
  for (int i = 0; i < adj[node].size(); i++) {
    if (!vis[adj[node][i]]) {
      vis[adj[node][i]] = true;
      int temp = dfs(adj, vis, adj[node][i]);
      if (temp == -1) {
        result = -1;
        break;
      } else if (temp == 0) {
        if (result == 0) {
          result = -1;
          break;
        } else {
          result = 1;
        }
      } else {
        if (result == 1) {
          result = -1;
          break;
        } else {
          result = 0;
        }
      }
    }
  }
  if (result == -2) {
    result = 0;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 0; i < n - 1; i++) {
    int t1, t2;
    cin >> t1 >> t2;
    adj[t1 - 1].push_back(t2 - 1);
    adj[t2 - 1].push_back(t1 - 1);
  }
  vector<bool> vis(n, false);
  int root = 0;
  while (adj[root].size() == 1) root++;
  vis[root] = true;
  int result = dfs(adj, vis, root);
  if (result == -1) {
    cout << 3 << " ";
  } else {
    cout << 1 << " ";
  }
  int leaves = 0;
  set<int> s;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaves += 1;
      s.insert(adj[i][0]);
    }
  }
  cout << (n - 1) - leaves + s.size() << '\n';
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 dfs(vector<vector<int>>& adj, vector<bool>& vis, int node) {
  int result = -2;
  for (int i = 0; i < adj[node].size(); i++) {
    if (!vis[adj[node][i]]) {
      vis[adj[node][i]] = true;
      int temp = dfs(adj, vis, adj[node][i]);
      if (temp == -1) {
        result = -1;
        break;
      } else if (temp == 0) {
        if (result == 0) {
          result = -1;
          break;
        } else {
          result = 1;
        }
      } else {
        if (result == 1) {
          result = -1;
          break;
        } else {
          result = 0;
        }
      }
    }
  }
  if (result == -2) {
    result = 0;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 0; i < n - 1; i++) {
    int t1, t2;
    cin >> t1 >> t2;
    adj[t1 - 1].push_back(t2 - 1);
    adj[t2 - 1].push_back(t1 - 1);
  }
  vector<bool> vis(n, false);
  int root = 0;
  while (adj[root].size() == 1) root++;
  vis[root] = true;
  int result = dfs(adj, vis, root);
  if (result == -1) {
    cout << 3 << " ";
  } else {
    cout << 1 << " ";
  }
  int leaves = 0;
  set<int> s;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaves += 1;
      s.insert(adj[i][0]);
    }
  }
  cout << (n - 1) - leaves + s.size() << '\n';
}

```