#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
vector<int> v[100001];
set<int> leaves, leavesSize;
bool visited[100001] = {0};
int dfs(int r, int depth) {
  visited[r] = true;
  if (leaves.find(r) != leaves.end()) {
    leavesSize.insert(depth);
  }
  for (int i = 0; i < v[r].size(); ++i) {
    if (!visited[v[r][i]]) {
      dfs(v[r][i], depth + 1);
    }
  }
  return 0;
}
int main() {
  int n;
  int a, b;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  set<int> parent;
  int edge;
  for (int i = 0; i < n; ++i) {
    if (v[i].size() == 1) {
      edge = i;
      leaves.insert(i);
      parent.insert(v[i][0]);
    }
  }
  int max = n - 1 - (leaves.size() - parent.size());
  dfs(edge, 0);
  int min = 1;
  for (auto e : leavesSize) {
    if (e % 2 == 1) {
      min = 3;
      break;
    }
  }
  cout << min << " " << max;
  return 0;
}

### 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;
using lli = long long int;
vector<int> v[100001];
set<int> leaves, leavesSize;
bool visited[100001] = {0};
int dfs(int r, int depth) {
  visited[r] = true;
  if (leaves.find(r) != leaves.end()) {
    leavesSize.insert(depth);
  }
  for (int i = 0; i < v[r].size(); ++i) {
    if (!visited[v[r][i]]) {
      dfs(v[r][i], depth + 1);
    }
  }
  return 0;
}
int main() {
  int n;
  int a, b;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  set<int> parent;
  int edge;
  for (int i = 0; i < n; ++i) {
    if (v[i].size() == 1) {
      edge = i;
      leaves.insert(i);
      parent.insert(v[i][0]);
    }
  }
  int max = n - 1 - (leaves.size() - parent.size());
  dfs(edge, 0);
  int min = 1;
  for (auto e : leavesSize) {
    if (e % 2 == 1) {
      min = 3;
      break;
    }
  }
  cout << min << " " << max;
  return 0;
}

```