#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
const long long int mod = 1e18;
int deg[100005];
vector<int> v[100005];
bool color[2] = {0};
map<int, int> ln, l;
void dfs(int x, int c, int par = -1) {
  if (v[x].size() == 1) {
    color[c] = true;
    l[x]++;
    if (par != -1) ln[par]++;
  }
  for (auto it : v[x]) {
    if (it != par) {
      dfs(it, c ^ 1, x);
    }
  }
}
int main() {
  int n, x, y;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      dfs(i, 1);
      break;
    }
  }
  if (color[0] && color[1]) {
    cout << 3 << " ";
  } else {
    cout << 1 << " ";
  }
  cout << n - 1 - (int)l.size() + (int)ln.size() << '\n';
}

### Prompt

In Cpp, your task is to solve 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;
using namespace std::chrono;
const long long int mod = 1e18;
int deg[100005];
vector<int> v[100005];
bool color[2] = {0};
map<int, int> ln, l;
void dfs(int x, int c, int par = -1) {
  if (v[x].size() == 1) {
    color[c] = true;
    l[x]++;
    if (par != -1) ln[par]++;
  }
  for (auto it : v[x]) {
    if (it != par) {
      dfs(it, c ^ 1, x);
    }
  }
}
int main() {
  int n, x, y;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      dfs(i, 1);
      break;
    }
  }
  if (color[0] && color[1]) {
    cout << 3 << " ";
  } else {
    cout << 1 << " ";
  }
  cout << n - 1 - (int)l.size() + (int)ln.size() << '\n';
}

```