#include <bits/stdc++.h>
using namespace std;
int n, root, mx, mn;
vector<int> tree[100005];
int dfs(int c, int p) {
  int e = 0, o = 0, tot = 0;
  for (auto& a : tree[c]) {
    if (a != p) {
      int t = dfs(a, c);
      o += t % 2;
      e += (t + 1) % 2;
      tot = t + 1;
    }
  }
  if (o && e) mn = 3;
  if (tree[c].size() == 1) return 0;
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    tree[a - 1].push_back(b - 1);
    tree[b - 1].push_back(a - 1);
  }
  for (int i = 0; i < n; i++)
    if (tree[i].size() - 1) root = i;
  mn = 1;
  mx = n - 1;
  for (int i = 0; i < n; i++) {
    int leaf = 0;
    for (auto& a : tree[i])
      if (tree[a].size() == 1) leaf++;
    mx -= max(leaf - 1, 0);
  }
  dfs(root, root);
  cout << mn << " " << mx << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int n, root, mx, mn;
vector<int> tree[100005];
int dfs(int c, int p) {
  int e = 0, o = 0, tot = 0;
  for (auto& a : tree[c]) {
    if (a != p) {
      int t = dfs(a, c);
      o += t % 2;
      e += (t + 1) % 2;
      tot = t + 1;
    }
  }
  if (o && e) mn = 3;
  if (tree[c].size() == 1) return 0;
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    tree[a - 1].push_back(b - 1);
    tree[b - 1].push_back(a - 1);
  }
  for (int i = 0; i < n; i++)
    if (tree[i].size() - 1) root = i;
  mn = 1;
  mx = n - 1;
  for (int i = 0; i < n; i++) {
    int leaf = 0;
    for (auto& a : tree[i])
      if (tree[a].size() == 1) leaf++;
    mx -= max(leaf - 1, 0);
  }
  dfs(root, root);
  cout << mn << " " << mx << endl;
}

```