#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    vector<vector<long long> > adj(n + 1);
    long long r = -1;
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
      if (adj[a].size() >= 2) {
        r = a;
      }
      if (adj[b].size() >= 2) {
        r = b;
      }
    }
    long long mx = n - 1, mn = INT_MAX;
    queue<pair<long long, long long> > q;
    q.push({r, 0});
    vector<long long> l;
    vector<long long> vis(n + 1, 0);
    vis[r] = 1;
    while (!q.empty()) {
      pair<long long, long long> t;
      t = q.front();
      q.pop();
      long long c = 0;
      for (auto i : adj[t.first]) {
        if (adj[i].size() == 1) {
          l.push_back(t.second + 1);
          c++;
        } else if (!vis[i])
          q.push({i, t.second + 1}), vis[i] = 1;
      }
      if (c > 0) mx -= (c - 1);
    }
    long long f1 = 0, f2 = 0;
    for (long long i = 0; i < l.size(); i++) {
      if (l[i] % 2)
        f1 = 1;
      else
        f2 = 1;
    }
    if (f1 && f2)
      mn = 3;
    else
      mn = 1;
    cout << mn << " " << mx;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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);
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    vector<vector<long long> > adj(n + 1);
    long long r = -1;
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
      if (adj[a].size() >= 2) {
        r = a;
      }
      if (adj[b].size() >= 2) {
        r = b;
      }
    }
    long long mx = n - 1, mn = INT_MAX;
    queue<pair<long long, long long> > q;
    q.push({r, 0});
    vector<long long> l;
    vector<long long> vis(n + 1, 0);
    vis[r] = 1;
    while (!q.empty()) {
      pair<long long, long long> t;
      t = q.front();
      q.pop();
      long long c = 0;
      for (auto i : adj[t.first]) {
        if (adj[i].size() == 1) {
          l.push_back(t.second + 1);
          c++;
        } else if (!vis[i])
          q.push({i, t.second + 1}), vis[i] = 1;
      }
      if (c > 0) mx -= (c - 1);
    }
    long long f1 = 0, f2 = 0;
    for (long long i = 0; i < l.size(); i++) {
      if (l[i] % 2)
        f1 = 1;
      else
        f2 = 1;
    }
    if (f1 && f2)
      mn = 3;
    else
      mn = 1;
    cout << mn << " " << mx;
  }
  return 0;
}

```