#include <bits/stdc++.h>
using namespace std;
using ll = long long;
mt19937 mt(736);
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  vector<vector<int>> gr(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  int mx = gr.size() - 1;
  for (const auto &row : gr) {
    int cnt = 0;
    for (auto it : row)
      if (gr[it].size() == 1) cnt++;
    mx -= max(0, cnt - 1);
  }
  vector<bool> col(gr.size());
  function<void(int, int)> dfs = [&gr, &col, &dfs](int v, int par) {
    if (par != -1) col[v] = !col[par];
    for (auto it : gr[v])
      if (it != par) dfs(it, v);
  };
  dfs(0, -1);
  vector<bool> types;
  int mn = 1;
  for (int i = 0; i < gr.size(); i++)
    if (gr[i].size() == 1) {
      types.push_back(col[i]);
      if (types.back() != types.front()) mn = 3;
    }
  cout << mn << ' ' << mx << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = long long;
mt19937 mt(736);
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  vector<vector<int>> gr(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  int mx = gr.size() - 1;
  for (const auto &row : gr) {
    int cnt = 0;
    for (auto it : row)
      if (gr[it].size() == 1) cnt++;
    mx -= max(0, cnt - 1);
  }
  vector<bool> col(gr.size());
  function<void(int, int)> dfs = [&gr, &col, &dfs](int v, int par) {
    if (par != -1) col[v] = !col[par];
    for (auto it : gr[v])
      if (it != par) dfs(it, v);
  };
  dfs(0, -1);
  vector<bool> types;
  int mn = 1;
  for (int i = 0; i < gr.size(); i++)
    if (gr[i].size() == 1) {
      types.push_back(col[i]);
      if (types.back() != types.front()) mn = 3;
    }
  cout << mn << ' ' << mx << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

```