#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g;
vector<int> deeps, p;
vector<bool> used;
set<int> leaves_parents;
void dfs(int v, int d = 1) {
  if (!used[v]) {
    used[v] = true;
    if (g[v].size() == 1) {
      deeps.push_back(d);
      leaves_parents.insert(g[v][0]);
    } else {
      for (auto c : g[v]) dfs(c, d + 1);
    }
  }
}
int main() {
  int n;
  cin >> n;
  g.resize(n + 1);
  p.resize(n + 1);
  used.resize(n + 1);
  for (int i = 1; i < n; ++i) {
    int x1, x2;
    cin >> x1 >> x2;
    g[x1].push_back(x2);
    g[x2].push_back(x1);
  }
  int root = 1;
  while (g[root].size() == 1) ++root;
  dfs(root);
  int i = 1;
  while (i < deeps.size() && deeps[i] % 2 == deeps[0] % 2) ++i;
  if (i == deeps.size())
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  int ans = n - 1;
  for (auto c : leaves_parents) {
    int quantity_leaves = 0;
    for (auto v : g[c]) quantity_leaves += (g[v].size() == 1);
    ans -= quantity_leaves - 1;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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<vector<int>> g;
vector<int> deeps, p;
vector<bool> used;
set<int> leaves_parents;
void dfs(int v, int d = 1) {
  if (!used[v]) {
    used[v] = true;
    if (g[v].size() == 1) {
      deeps.push_back(d);
      leaves_parents.insert(g[v][0]);
    } else {
      for (auto c : g[v]) dfs(c, d + 1);
    }
  }
}
int main() {
  int n;
  cin >> n;
  g.resize(n + 1);
  p.resize(n + 1);
  used.resize(n + 1);
  for (int i = 1; i < n; ++i) {
    int x1, x2;
    cin >> x1 >> x2;
    g[x1].push_back(x2);
    g[x2].push_back(x1);
  }
  int root = 1;
  while (g[root].size() == 1) ++root;
  dfs(root);
  int i = 1;
  while (i < deeps.size() && deeps[i] % 2 == deeps[0] % 2) ++i;
  if (i == deeps.size())
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  int ans = n - 1;
  for (auto c : leaves_parents) {
    int quantity_leaves = 0;
    for (auto v : g[c]) quantity_leaves += (g[v].size() == 1);
    ans -= quantity_leaves - 1;
  }
  cout << ans;
  return 0;
}

```