#include <bits/stdc++.h>
using namespace std;
int visited[100001];
vector<int> graph[100001];
int flag = 0;
int leaf[100001];
int depth[100001];
int direct_leaf_parent[100001];
void add_edge(int x, int y) {
  graph[x].push_back(y);
  graph[y].push_back(x);
}
void dfs(int u, int d) {
  visited[u] = 1;
  depth[u] = d;
  if (graph[u].size() == 1) leaf[u] = 1;
  for (int i : graph[u]) {
    if (graph[i].size() == 1) direct_leaf_parent[u] += 1;
    if (!visited[i]) dfs(i, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    add_edge(x, y);
  }
  dfs(1, 1);
  vector<int> v;
  int min_ans = 0, max_ans = n - 1;
  int odd = 0, even = 0, count = 0;
  for (int i = 1; i <= n; i++) {
    if (leaf[i] == 1) {
      if (depth[i] % 2 == 0)
        even = 1;
      else
        odd = 1;
    }
    if (direct_leaf_parent[i] > 1) {
      max_ans -= direct_leaf_parent[i];
      max_ans += 1;
    }
  }
  if (odd == 1 && even == 1)
    min_ans = 3;
  else
    min_ans = 1;
  cout << min_ans << " " << max_ans << "\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 visited[100001];
vector<int> graph[100001];
int flag = 0;
int leaf[100001];
int depth[100001];
int direct_leaf_parent[100001];
void add_edge(int x, int y) {
  graph[x].push_back(y);
  graph[y].push_back(x);
}
void dfs(int u, int d) {
  visited[u] = 1;
  depth[u] = d;
  if (graph[u].size() == 1) leaf[u] = 1;
  for (int i : graph[u]) {
    if (graph[i].size() == 1) direct_leaf_parent[u] += 1;
    if (!visited[i]) dfs(i, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    add_edge(x, y);
  }
  dfs(1, 1);
  vector<int> v;
  int min_ans = 0, max_ans = n - 1;
  int odd = 0, even = 0, count = 0;
  for (int i = 1; i <= n; i++) {
    if (leaf[i] == 1) {
      if (depth[i] % 2 == 0)
        even = 1;
      else
        odd = 1;
    }
    if (direct_leaf_parent[i] > 1) {
      max_ans -= direct_leaf_parent[i];
      max_ans += 1;
    }
  }
  if (odd == 1 && even == 1)
    min_ans = 3;
  else
    min_ans = 1;
  cout << min_ans << " " << max_ans << "\n";
}

```