#include <bits/stdc++.h>
using namespace std;
int n;
const int maxn = 1e5 + 1;
int deg[maxn], odd[maxn], even[maxn], v[maxn], cnt[maxn];
vector<vector<int>> e(maxn, vector<int>());
void dfs(int node) {
  if (deg[node] == 1) {
    even[node] = 1;
    return;
  }
  v[node] = 1;
  for (int i = 0; i < e[node].size(); ++i) {
    int child = e[node][i];
    if (!v[child]) {
      dfs(child);
      if (even[child]) odd[node] = 1;
      if (odd[child]) even[node] = 1;
    }
  }
}
void find() {
  int res1 = 1;
  int res2 = n - 1;
  for (int i = 0; i < n; ++i)
    if (odd[i] && even[i]) res1 = 3;
  for (int i = 0; i < n; ++i)
    if (deg[i] == 1) cnt[e[i][0]]++;
  for (int i = 0; i < n; ++i)
    if (cnt[i] > 0) res2 -= cnt[i] - 1;
  cout << res1 << " " << res2 << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    deg[a]++;
    deg[b]++;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  for (int i = 0; i < n; ++i)
    if (deg[i] != 1) {
      dfs(i);
      break;
    }
  find();
  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;
int n;
const int maxn = 1e5 + 1;
int deg[maxn], odd[maxn], even[maxn], v[maxn], cnt[maxn];
vector<vector<int>> e(maxn, vector<int>());
void dfs(int node) {
  if (deg[node] == 1) {
    even[node] = 1;
    return;
  }
  v[node] = 1;
  for (int i = 0; i < e[node].size(); ++i) {
    int child = e[node][i];
    if (!v[child]) {
      dfs(child);
      if (even[child]) odd[node] = 1;
      if (odd[child]) even[node] = 1;
    }
  }
}
void find() {
  int res1 = 1;
  int res2 = n - 1;
  for (int i = 0; i < n; ++i)
    if (odd[i] && even[i]) res1 = 3;
  for (int i = 0; i < n; ++i)
    if (deg[i] == 1) cnt[e[i][0]]++;
  for (int i = 0; i < n; ++i)
    if (cnt[i] > 0) res2 -= cnt[i] - 1;
  cout << res1 << " " << res2 << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    deg[a]++;
    deg[b]++;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  for (int i = 0; i < n; ++i)
    if (deg[i] != 1) {
      dfs(i);
      break;
    }
  find();
  return 0;
}

```