#include <bits/stdc++.h>
using namespace std;
vector<long long> adj[100005];
long long vis[100005], mn, mx, odd[100005], even[100005], leaf[100005];
void dfs(long long src) {
  vis[src] = 1;
  leaf[src] = 1;
  long long cnt = 0;
  for (long long i = 0; i < adj[src].size(); i++) {
    long long node = adj[src][i];
    if (vis[node] == 0) {
      dfs(node);
      leaf[src] = 0;
      cnt += leaf[node];
      odd[src] |= even[node];
      even[src] |= odd[node];
    }
  }
  mx -= max((long long)0, (cnt - 1));
  if (odd[src] && even[src]) mn = 3;
  if (leaf[src]) even[src] = 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long root = 1;
  for (long long i = 1; i <= n; i++) {
    long long sz = adj[i].size();
    if (sz != 1) {
      root = i;
      break;
    }
  }
  mx = n - 1;
  mn = 1;
  dfs(root);
  cout << mn << " " << mx << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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<long long> adj[100005];
long long vis[100005], mn, mx, odd[100005], even[100005], leaf[100005];
void dfs(long long src) {
  vis[src] = 1;
  leaf[src] = 1;
  long long cnt = 0;
  for (long long i = 0; i < adj[src].size(); i++) {
    long long node = adj[src][i];
    if (vis[node] == 0) {
      dfs(node);
      leaf[src] = 0;
      cnt += leaf[node];
      odd[src] |= even[node];
      even[src] |= odd[node];
    }
  }
  mx -= max((long long)0, (cnt - 1));
  if (odd[src] && even[src]) mn = 3;
  if (leaf[src]) even[src] = 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long root = 1;
  for (long long i = 1; i <= n; i++) {
    long long sz = adj[i].size();
    if (sz != 1) {
      root = i;
      break;
    }
  }
  mx = n - 1;
  mn = 1;
  dfs(root);
  cout << mn << " " << mx << "\n";
  return 0;
}

```