#include <bits/stdc++.h>
using namespace std;
long long mod = (long long)1e9 + 7;
const long long N = 10000000 + 9;
set<long long> dv;
vector<vector<long long> > gr;
long long l, m;
bool is(long long u) { return gr[u].size() == 1; }
void dfs(long long u, long long par, long long h = 0) {
  if (gr[u].size() == 1) dv.insert(h % 2), l++;
  bool flg = 0;
  for (auto i : gr[u]) {
    if (is(i)) flg = 1;
    if (i != par) dfs(i, u, h + 1);
  }
  if (flg) m++;
}
void Solve() {
  long long n;
  cin >> n;
  gr.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  dfs(1, -1);
  if (dv.size() == 1)
    cout << "1 ";
  else
    cout << "3 ";
  cout << (n - 1) - l + m << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1, tc = 1;
  while (t--) {
    Solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long mod = (long long)1e9 + 7;
const long long N = 10000000 + 9;
set<long long> dv;
vector<vector<long long> > gr;
long long l, m;
bool is(long long u) { return gr[u].size() == 1; }
void dfs(long long u, long long par, long long h = 0) {
  if (gr[u].size() == 1) dv.insert(h % 2), l++;
  bool flg = 0;
  for (auto i : gr[u]) {
    if (is(i)) flg = 1;
    if (i != par) dfs(i, u, h + 1);
  }
  if (flg) m++;
}
void Solve() {
  long long n;
  cin >> n;
  gr.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  dfs(1, -1);
  if (dv.size() == 1)
    cout << "1 ";
  else
    cout << "3 ";
  cout << (n - 1) - l + m << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1, tc = 1;
  while (t--) {
    Solve();
  }
  return 0;
}

```