#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > adj;
void DFS(long long v, long long p, long long dep, vector<long long> &depth) {
  depth[v] = dep;
  for (auto x : adj[v]) {
    if (x == p) continue;
    DFS(x, v, dep + 1, depth);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  adj.resize(n);
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<bool> leaf(n, false);
  for (long long i = 0; i < n; i++) {
    if (adj[i].size() == 1) leaf[i] = true;
  }
  vector<long long> depth(n);
  long long root = 0;
  for (; root < n; root++) {
    if (leaf[root]) {
      DFS(root, -1, 0, depth);
      break;
    }
  }
  long long min = 1;
  for (long long i = 0; i < n; i++) {
    if (leaf[i] && depth[i] % 2) min = 3;
  }
  long long max = n - 1;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    if (leaf[i]) {
      long long x = adj[i].front();
      mp[x]++;
    }
  }
  for (auto x : mp) {
    max -= x.second - 1;
  }
  cout << min << " " << max;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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> > adj;
void DFS(long long v, long long p, long long dep, vector<long long> &depth) {
  depth[v] = dep;
  for (auto x : adj[v]) {
    if (x == p) continue;
    DFS(x, v, dep + 1, depth);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  adj.resize(n);
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<bool> leaf(n, false);
  for (long long i = 0; i < n; i++) {
    if (adj[i].size() == 1) leaf[i] = true;
  }
  vector<long long> depth(n);
  long long root = 0;
  for (; root < n; root++) {
    if (leaf[root]) {
      DFS(root, -1, 0, depth);
      break;
    }
  }
  long long min = 1;
  for (long long i = 0; i < n; i++) {
    if (leaf[i] && depth[i] % 2) min = 3;
  }
  long long max = n - 1;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    if (leaf[i]) {
      long long x = adj[i].front();
      mp[x]++;
    }
  }
  for (auto x : mp) {
    max -= x.second - 1;
  }
  cout << min << " " << max;
  return 0;
}

```