#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 5;
vector<int> tree[MAX_N];
vector<int> depth(MAX_N, 0);
vector<int> no_of_leafs(MAX_N, 0);
vector<int> parent(MAX_N, 0);
vector<int> is_leaf(MAX_N, false);
void dfs_depth(int v, int parent_v) {
  if (depth[v] != 0) return;
  parent[v] = parent_v;
  depth[v] = depth[parent_v] + 1;
  no_of_leafs[v] = (tree[parent_v].size() == 1 ? 1 : 0);
  for (int i = 0; i < tree[v].size(); i++) {
    int child_v = tree[v][i];
    if (child_v == parent_v) {
      continue;
    }
    dfs_depth(child_v, v);
    no_of_leafs[v] += (tree[child_v].size() == 1);
  }
}
int main() {
  int no_of_vertices;
  cin >> no_of_vertices;
  int no_of_edges = no_of_vertices - 1;
  for (int i = 1; i <= no_of_edges; i++) {
    int u, v;
    cin >> u >> v;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  for (int i = 1; i <= no_of_vertices; i++) {
    if (tree[i].size() == 1) {
      is_leaf[i] = true;
    }
  }
  dfs_depth(1, 0);
  set<int> leaf_depth_parities;
  for (int i = 1; i <= no_of_vertices; i++) {
    if (is_leaf[i]) {
      leaf_depth_parities.insert(depth[i] % 2);
    }
  }
  int minimum_colours = (leaf_depth_parities.size() == 1 ? 1 : 3);
  int maximum_colours = no_of_edges;
  for (int v = 1; v <= no_of_vertices; v++) {
    if (no_of_leafs[v] >= 1) {
      maximum_colours++;
    }
    if (is_leaf[v]) {
      maximum_colours--;
    }
  }
  cout << minimum_colours << " " << maximum_colours << "\n";
  return 0;
}

### Prompt

Generate 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;
const int MAX_N = 1e5 + 5;
vector<int> tree[MAX_N];
vector<int> depth(MAX_N, 0);
vector<int> no_of_leafs(MAX_N, 0);
vector<int> parent(MAX_N, 0);
vector<int> is_leaf(MAX_N, false);
void dfs_depth(int v, int parent_v) {
  if (depth[v] != 0) return;
  parent[v] = parent_v;
  depth[v] = depth[parent_v] + 1;
  no_of_leafs[v] = (tree[parent_v].size() == 1 ? 1 : 0);
  for (int i = 0; i < tree[v].size(); i++) {
    int child_v = tree[v][i];
    if (child_v == parent_v) {
      continue;
    }
    dfs_depth(child_v, v);
    no_of_leafs[v] += (tree[child_v].size() == 1);
  }
}
int main() {
  int no_of_vertices;
  cin >> no_of_vertices;
  int no_of_edges = no_of_vertices - 1;
  for (int i = 1; i <= no_of_edges; i++) {
    int u, v;
    cin >> u >> v;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  for (int i = 1; i <= no_of_vertices; i++) {
    if (tree[i].size() == 1) {
      is_leaf[i] = true;
    }
  }
  dfs_depth(1, 0);
  set<int> leaf_depth_parities;
  for (int i = 1; i <= no_of_vertices; i++) {
    if (is_leaf[i]) {
      leaf_depth_parities.insert(depth[i] % 2);
    }
  }
  int minimum_colours = (leaf_depth_parities.size() == 1 ? 1 : 3);
  int maximum_colours = no_of_edges;
  for (int v = 1; v <= no_of_vertices; v++) {
    if (no_of_leafs[v] >= 1) {
      maximum_colours++;
    }
    if (is_leaf[v]) {
      maximum_colours--;
    }
  }
  cout << minimum_colours << " " << maximum_colours << "\n";
  return 0;
}

```