#include <bits/stdc++.h>
using namespace std;
vector<vector<long long int>> adj;
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int com_leaves = 0;
  long long int n;
  cin >> n;
  long long int a, b;
  adj.resize(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<long long int> dist(n + 1, 9223372036854775807);
  vector<long long int> visi(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1 && visi[i] == 0) {
      visi[i] = 1;
      long long int cnt = 1;
      for (auto x : adj[adj[i][0]]) {
        if (x == i) continue;
        if (adj[x].size() == 1) {
          visi[x] = 1;
          cnt++;
        }
      }
      com_leaves += cnt - 1;
    }
  }
  long long int source = 1;
  for (int i = 1; i <= n; i++) {
    if (visi[i] == 1) source = i;
  }
  queue<long long int> que;
  que.push(source);
  dist[source] = 0;
  while (!que.empty()) {
    long long int x = que.front();
    que.pop();
    for (auto k : adj[x]) {
      if (dist[k] == 9223372036854775807) {
        dist[k] = dist[x] + 1;
        que.push(k);
      }
    }
  }
  long long int fl = 0;
  for (int i = 1; i <= n; i++) {
    if (visi[i] == 1) {
      if (dist[i] % 2 == 1) fl = 1;
    }
  }
  if (fl == 0) {
    cout << 1 << " ";
  } else
    cout << 3 << " ";
  cout << n - 1 - com_leaves << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
vector<vector<long long int>> adj;
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int com_leaves = 0;
  long long int n;
  cin >> n;
  long long int a, b;
  adj.resize(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<long long int> dist(n + 1, 9223372036854775807);
  vector<long long int> visi(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1 && visi[i] == 0) {
      visi[i] = 1;
      long long int cnt = 1;
      for (auto x : adj[adj[i][0]]) {
        if (x == i) continue;
        if (adj[x].size() == 1) {
          visi[x] = 1;
          cnt++;
        }
      }
      com_leaves += cnt - 1;
    }
  }
  long long int source = 1;
  for (int i = 1; i <= n; i++) {
    if (visi[i] == 1) source = i;
  }
  queue<long long int> que;
  que.push(source);
  dist[source] = 0;
  while (!que.empty()) {
    long long int x = que.front();
    que.pop();
    for (auto k : adj[x]) {
      if (dist[k] == 9223372036854775807) {
        dist[k] = dist[x] + 1;
        que.push(k);
      }
    }
  }
  long long int fl = 0;
  for (int i = 1; i <= n; i++) {
    if (visi[i] == 1) {
      if (dist[i] % 2 == 1) fl = 1;
    }
  }
  if (fl == 0) {
    cout << 1 << " ";
  } else
    cout << 3 << " ";
  cout << n - 1 - com_leaves << '\n';
  return 0;
}

```