#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
const ll MOD = 1e9 + 7;
const ll inf = (LLONG_MAX >> 2);
bool dfs1(int node, int pere, vector<vector<int>>& adj, vector<int>& deg,
          int l) {
  for (int i : adj[node]) {
    if (i == pere) continue;
    if (deg[i] == 1) {
      if ((l & 1) == 0) return true;
    } else if (dfs1(i, node, adj, deg, l + 1))
      return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  vector<int> deg(n + 1);
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    ++deg[a];
    ++deg[b];
  }
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      int l = 0;
      if (dfs1(i, -1, adj, deg, l))
        cout << "3 ";
      else
        cout << "1 ";
      break;
    }
  }
  int ans = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] > 1) {
      int cnt = 0;
      for (int j : adj[i]) {
        if (deg[j] == 1) ++cnt;
      }
      if (cnt > 1) ans -= (cnt - 1);
    }
  }
  cout << ans;
  return 0;
}

### 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;
using ll = int64_t;
const ll MOD = 1e9 + 7;
const ll inf = (LLONG_MAX >> 2);
bool dfs1(int node, int pere, vector<vector<int>>& adj, vector<int>& deg,
          int l) {
  for (int i : adj[node]) {
    if (i == pere) continue;
    if (deg[i] == 1) {
      if ((l & 1) == 0) return true;
    } else if (dfs1(i, node, adj, deg, l + 1))
      return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  vector<int> deg(n + 1);
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    ++deg[a];
    ++deg[b];
  }
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      int l = 0;
      if (dfs1(i, -1, adj, deg, l))
        cout << "3 ";
      else
        cout << "1 ";
      break;
    }
  }
  int ans = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] > 1) {
      int cnt = 0;
      for (int j : adj[i]) {
        if (deg[j] == 1) ++cnt;
      }
      if (cnt > 1) ans -= (cnt - 1);
    }
  }
  cout << ans;
  return 0;
}

```