#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> ad(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    ad[a].push_back(b);
    ad[b].push_back(a);
  }
  int root = -1;
  for (int i = 0; i < n; i++) {
    if ((int)ad[i].size() == 1) root = i;
  }
  vector<int> dists(n);
  function<void(int, int, int)> dfs0 = [&](int cur, int par, int d) {
    dists[cur] = d;
    for (int ch : ad[cur]) {
      if (ch == par) continue;
      dfs0(ch, cur, d + 1);
    }
  };
  dfs0(root, -1, 0);
  bool bo = true;
  for (int i = 0; i < n; i++) {
    if ((int)ad[i].size() > 1) continue;
    if (dists[i] & 1) bo = false;
  }
  if (bo) {
    cout << 1 << " ";
  } else {
    cout << 3 << " ";
  }
  int res = 0;
  function<void(int, int)> dfs1 = [&](int cur, int par) {
    bool bo2 = false;
    for (int ch : ad[cur]) {
      if (ch == par) continue;
      if (ad[ch].size() == 1) {
        if (dists[ch] != 2) bo2 = true;
      } else {
        res++;
        dfs1(ch, cur);
      }
    }
    if (bo2) res++;
  };
  dfs1(root, -1);
  cout << res << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> ad(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    ad[a].push_back(b);
    ad[b].push_back(a);
  }
  int root = -1;
  for (int i = 0; i < n; i++) {
    if ((int)ad[i].size() == 1) root = i;
  }
  vector<int> dists(n);
  function<void(int, int, int)> dfs0 = [&](int cur, int par, int d) {
    dists[cur] = d;
    for (int ch : ad[cur]) {
      if (ch == par) continue;
      dfs0(ch, cur, d + 1);
    }
  };
  dfs0(root, -1, 0);
  bool bo = true;
  for (int i = 0; i < n; i++) {
    if ((int)ad[i].size() > 1) continue;
    if (dists[i] & 1) bo = false;
  }
  if (bo) {
    cout << 1 << " ";
  } else {
    cout << 3 << " ";
  }
  int res = 0;
  function<void(int, int)> dfs1 = [&](int cur, int par) {
    bool bo2 = false;
    for (int ch : ad[cur]) {
      if (ch == par) continue;
      if (ad[ch].size() == 1) {
        if (dists[ch] != 2) bo2 = true;
      } else {
        res++;
        dfs1(ch, cur);
      }
    }
    if (bo2) res++;
  };
  dfs1(root, -1);
  cout << res << "\n";
  return 0;
}

```