#include <bits/stdc++.h>
using namespace std;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int MAXN = 1e5 + 5;
vector<int> G[MAXN];
int small = 1, wasted;
bool is_leaf(const int &u) { return (int)G[u].size() == 1; }
void dfs(int u, int parent, bool depth) {
  if (is_leaf(u) && depth) small = 3;
  int children_leaves = is_leaf(parent);
  for (const int &v : G[u])
    if (v != parent) {
      dfs(v, u, depth ^ 1);
      children_leaves += is_leaf(v);
    }
  if (children_leaves > 1) wasted += children_leaves - 1;
}
void solve() {
  int N;
  cin >> N;
  for (int i = 1; i < N; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  for (int u = 1; u <= N; ++u)
    if (is_leaf(u)) {
      dfs(u, 0, false);
      break;
    }
  cout << small << ' ' << N - 1 - wasted << '\n';
}
int main() {
  fastIO();
  solve();
  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;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int MAXN = 1e5 + 5;
vector<int> G[MAXN];
int small = 1, wasted;
bool is_leaf(const int &u) { return (int)G[u].size() == 1; }
void dfs(int u, int parent, bool depth) {
  if (is_leaf(u) && depth) small = 3;
  int children_leaves = is_leaf(parent);
  for (const int &v : G[u])
    if (v != parent) {
      dfs(v, u, depth ^ 1);
      children_leaves += is_leaf(v);
    }
  if (children_leaves > 1) wasted += children_leaves - 1;
}
void solve() {
  int N;
  cin >> N;
  for (int i = 1; i < N; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  for (int u = 1; u <= N; ++u)
    if (is_leaf(u)) {
      dfs(u, 0, false);
      break;
    }
  cout << small << ' ' << N - 1 - wasted << '\n';
}
int main() {
  fastIO();
  solve();
  return 0;
}

```