#include <bits/stdc++.h>
using namespace std;
unordered_map<int, vector<int>> nodes;
unordered_map<int, int> visited;
set<int> x;
void dfs(int s, int p, int k) {
  visited[s] = 1;
  bool ok = 1;
  for (auto i : nodes[s]) {
    if (visited.count(i) == 0) {
      ok = 0;
      dfs(i, s, k + 1);
    }
  }
  if (ok) {
    x.insert(k % 2);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    nodes[a].push_back(b);
    nodes[b].push_back(a);
  }
  dfs(1, 0, 0);
  int c = n - 1;
  unordered_map<int, int> dp;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    if (nodes[i].size() == 1) {
      k++;
      dp[nodes[i][0]]++;
    }
  }
  for (auto i : dp) {
    c -= (i.second - 1);
  }
  if (k == 2 && n % 2) {
    cout << 1 << ' ';
  } else if (k == 2) {
    cout << 3 << ' ';
  } else if (x.size() == 1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << c << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
unordered_map<int, vector<int>> nodes;
unordered_map<int, int> visited;
set<int> x;
void dfs(int s, int p, int k) {
  visited[s] = 1;
  bool ok = 1;
  for (auto i : nodes[s]) {
    if (visited.count(i) == 0) {
      ok = 0;
      dfs(i, s, k + 1);
    }
  }
  if (ok) {
    x.insert(k % 2);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    nodes[a].push_back(b);
    nodes[b].push_back(a);
  }
  dfs(1, 0, 0);
  int c = n - 1;
  unordered_map<int, int> dp;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    if (nodes[i].size() == 1) {
      k++;
      dp[nodes[i][0]]++;
    }
  }
  for (auto i : dp) {
    c -= (i.second - 1);
  }
  if (k == 2 && n % 2) {
    cout << 1 << ' ';
  } else if (k == 2) {
    cout << 3 << ' ';
  } else if (x.size() == 1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << c << '\n';
  return 0;
}

```