#include <bits/stdc++.h>
using namespace std;
void dfs(vector<vector<int>> &adj, vector<bool> &leaf,
         vector<bool> &leaf_connected, vector<bool> &visited,
         vector<long long> &distanc, int u, int root) {
  visited[u] = 1;
  for (int v : adj[u]) {
    if (root == u) leaf_connected[v] = 1;
    if (!visited[v]) {
      distanc[v] += distanc[u] + 1;
      if (leaf[v])
        leaf_connected[u] = 1;
      else
        dfs(adj, leaf, leaf_connected, visited, distanc, v, root);
    }
  }
}
pair<long long, long long> ansi(vector<vector<int>> &adj, vector<bool> &leaf,
                                int n, int root) {
  vector<long long> distanc(n + 1, 1e9);
  pair<long long, long long> ans;
  vector<bool> leaf_connected(n + 1, 0);
  vector<bool> visited(n + 1, 0);
  distanc[root] = 0;
  dfs(adj, leaf, leaf_connected, visited, distanc, root, root);
  bool prev_parity = 0;
  int i = 1, leaf_count = 0, leaf_connector = 0;
  for (; i <= n; i++) {
    if (leaf[i]) {
      prev_parity = distanc[i] & 1;
      i++;
      break;
    }
  }
  bool flag = 0;
  for (; i <= n; i++) {
    if (leaf[i] && (distanc[i] & 1) != prev_parity) flag = 1;
  }
  if (!flag) {
    ans.first = 1;
  } else
    ans.first = 3;
  for (int i = 1; i <= n; i++) {
    if (leaf[i])
      leaf_count++;
    else if (leaf_connected[i])
      leaf_connector++;
  }
  ans.second = n - 1 - leaf_count + leaf_connector;
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  vector<int> indegree(n + 1, 0);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    indegree[x]++;
    indegree[y]++;
  }
  vector<bool> leaf(n + 1, false);
  int root;
  for (int i = 1; i <= n; i++)
    if (indegree[i] == 1) leaf[i] = 1, root = i;
  pair<long long, long long> ans = ansi(adj, leaf, n, root);
  cout << ans.first << " " << ans.second << endl;
  return 0;
}

### Prompt

Please formulate 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;
void dfs(vector<vector<int>> &adj, vector<bool> &leaf,
         vector<bool> &leaf_connected, vector<bool> &visited,
         vector<long long> &distanc, int u, int root) {
  visited[u] = 1;
  for (int v : adj[u]) {
    if (root == u) leaf_connected[v] = 1;
    if (!visited[v]) {
      distanc[v] += distanc[u] + 1;
      if (leaf[v])
        leaf_connected[u] = 1;
      else
        dfs(adj, leaf, leaf_connected, visited, distanc, v, root);
    }
  }
}
pair<long long, long long> ansi(vector<vector<int>> &adj, vector<bool> &leaf,
                                int n, int root) {
  vector<long long> distanc(n + 1, 1e9);
  pair<long long, long long> ans;
  vector<bool> leaf_connected(n + 1, 0);
  vector<bool> visited(n + 1, 0);
  distanc[root] = 0;
  dfs(adj, leaf, leaf_connected, visited, distanc, root, root);
  bool prev_parity = 0;
  int i = 1, leaf_count = 0, leaf_connector = 0;
  for (; i <= n; i++) {
    if (leaf[i]) {
      prev_parity = distanc[i] & 1;
      i++;
      break;
    }
  }
  bool flag = 0;
  for (; i <= n; i++) {
    if (leaf[i] && (distanc[i] & 1) != prev_parity) flag = 1;
  }
  if (!flag) {
    ans.first = 1;
  } else
    ans.first = 3;
  for (int i = 1; i <= n; i++) {
    if (leaf[i])
      leaf_count++;
    else if (leaf_connected[i])
      leaf_connector++;
  }
  ans.second = n - 1 - leaf_count + leaf_connector;
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  vector<int> indegree(n + 1, 0);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    indegree[x]++;
    indegree[y]++;
  }
  vector<bool> leaf(n + 1, false);
  int root;
  for (int i = 1; i <= n; i++)
    if (indegree[i] == 1) leaf[i] = 1, root = i;
  pair<long long, long long> ans = ansi(adj, leaf, n, root);
  cout << ans.first << " " << ans.second << endl;
  return 0;
}

```