#include <bits/stdc++.h>
using namespace std;
bool odd = false;
int path = 0, cnt1 = 0, cnt2 = 0;
void dfs(vector<int> v[], int i, int h, int p) {
  bool flag = true, f = false;
  for (auto it = v[i].begin(); it != v[i].end(); it++) {
    if (*it != p) {
      dfs(v, *it, h + 1, i);
      flag = false;
    }
    if (v[*it].size() == 1) f = true;
  }
  if (f) cnt2++;
  if (flag && h % 2) odd = true;
  if (flag) path = max(path, h), cnt1++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t, n, i, x, y;
  cin >> n;
  vector<int> v[n + 1];
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int node = 1;
  for (i = 1; i < n + 1; i++)
    if (v[i].size() == 1) node = i;
  dfs(v, node, 0, 0);
  cnt1++;
  if (odd)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << n - 1 - cnt1 + cnt2 << endl;
}

### Prompt

Please formulate 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;
bool odd = false;
int path = 0, cnt1 = 0, cnt2 = 0;
void dfs(vector<int> v[], int i, int h, int p) {
  bool flag = true, f = false;
  for (auto it = v[i].begin(); it != v[i].end(); it++) {
    if (*it != p) {
      dfs(v, *it, h + 1, i);
      flag = false;
    }
    if (v[*it].size() == 1) f = true;
  }
  if (f) cnt2++;
  if (flag && h % 2) odd = true;
  if (flag) path = max(path, h), cnt1++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t, n, i, x, y;
  cin >> n;
  vector<int> v[n + 1];
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int node = 1;
  for (i = 1; i < n + 1; i++)
    if (v[i].size() == 1) node = i;
  dfs(v, node, 0, 0);
  cnt1++;
  if (odd)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << n - 1 - cnt1 + cnt2 << endl;
}

```