#include <bits/stdc++.h>
using namespace std;
vector<long long> vis;
long long odd_flag;
long long dia = -1;
long long leaves_next_to_each_other = 0;
vector<vector<long long>> adj;
long long foo(long long root) {
  vis[root] = 1;
  vector<long long> h;
  vector<long long> oe(2, 0);
  long long sum = 0;
  for (auto a : adj[root]) {
    if (!vis[a]) {
      long long tmp = foo(a);
      h.push_back(tmp);
      if (oe[(tmp + 1) % 2]) odd_flag = 1;
      oe[tmp % 2]++;
      sum += tmp;
    }
  }
  long long i;
  sort(h.begin(), h.end());
  for (i = 0; i < h.size(); i++) {
    if (h[i] > 1) break;
  }
  if (i > 0) leaves_next_to_each_other += i - 1;
  if (h.size() == 0) return 1;
  if (h.size() > 1) dia = max(dia, sum);
  return h[h.size() - 1] + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  odd_flag = 0;
  for (long long i = 0; i < n; i++) vis.push_back(0);
  vector<long long> empt;
  for (long long i = 0; i < n; i++) adj.push_back(empt);
  for (long long i = 0; i < n - 1; i++) {
    long long tmp1, tmp2;
    cin >> tmp1 >> tmp2;
    tmp1--;
    tmp2--;
    adj[tmp1].push_back(tmp2);
    adj[tmp2].push_back(tmp1);
  }
  for (long long i = 0; i < n; i++)
    if (adj[i].size() > 1) {
      foo(i);
      break;
    }
  if (odd_flag)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << n - 1 - leaves_next_to_each_other << endl;
  return 0;
}

### Prompt

Your task is to create 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;
vector<long long> vis;
long long odd_flag;
long long dia = -1;
long long leaves_next_to_each_other = 0;
vector<vector<long long>> adj;
long long foo(long long root) {
  vis[root] = 1;
  vector<long long> h;
  vector<long long> oe(2, 0);
  long long sum = 0;
  for (auto a : adj[root]) {
    if (!vis[a]) {
      long long tmp = foo(a);
      h.push_back(tmp);
      if (oe[(tmp + 1) % 2]) odd_flag = 1;
      oe[tmp % 2]++;
      sum += tmp;
    }
  }
  long long i;
  sort(h.begin(), h.end());
  for (i = 0; i < h.size(); i++) {
    if (h[i] > 1) break;
  }
  if (i > 0) leaves_next_to_each_other += i - 1;
  if (h.size() == 0) return 1;
  if (h.size() > 1) dia = max(dia, sum);
  return h[h.size() - 1] + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  odd_flag = 0;
  for (long long i = 0; i < n; i++) vis.push_back(0);
  vector<long long> empt;
  for (long long i = 0; i < n; i++) adj.push_back(empt);
  for (long long i = 0; i < n - 1; i++) {
    long long tmp1, tmp2;
    cin >> tmp1 >> tmp2;
    tmp1--;
    tmp2--;
    adj[tmp1].push_back(tmp2);
    adj[tmp2].push_back(tmp1);
  }
  for (long long i = 0; i < n; i++)
    if (adj[i].size() > 1) {
      foo(i);
      break;
    }
  if (odd_flag)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << n - 1 - leaves_next_to_each_other << endl;
  return 0;
}

```