#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ";
  err(++it, args...);
}
vector<vector<long long int>> adj;
vector<long long int> color;
void dfs(long long int v, long long int root, long long int col) {
  color[v] = col;
  for (long long int x : adj[v]) {
    if (x == root) continue;
    dfs(x, v, 1 - col);
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  color = vector<long long int>(n);
  adj = vector<vector<long long int>>(n);
  for (long long int i = 0, a, b; i < n - 1; i++) {
    cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1, 0);
  long long int chk_even = 0;
  vector<bool> asNleaf(n, false);
  long long int leaf_cnt = 0;
  for (long long int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaf_cnt++;
      chk_even |= (1 << color[i]);
      asNleaf[adj[i][0]] = true;
    }
  }
  long long int ans_mn = 0, ans_mx = 0;
  if (chk_even != 3) {
    ans_mn = 1;
  } else {
    ans_mn = 3;
  }
  long long int cnt_adjL = 0;
  for (long long int i = 0; i < n; i++) {
    if (asNleaf[i]) ++cnt_adjL;
  }
  ans_mx = (n - 1) - leaf_cnt + cnt_adjL;
  cout << ans_mn << " " << ans_mx << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ";
  err(++it, args...);
}
vector<vector<long long int>> adj;
vector<long long int> color;
void dfs(long long int v, long long int root, long long int col) {
  color[v] = col;
  for (long long int x : adj[v]) {
    if (x == root) continue;
    dfs(x, v, 1 - col);
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  color = vector<long long int>(n);
  adj = vector<vector<long long int>>(n);
  for (long long int i = 0, a, b; i < n - 1; i++) {
    cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1, 0);
  long long int chk_even = 0;
  vector<bool> asNleaf(n, false);
  long long int leaf_cnt = 0;
  for (long long int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaf_cnt++;
      chk_even |= (1 << color[i]);
      asNleaf[adj[i][0]] = true;
    }
  }
  long long int ans_mn = 0, ans_mx = 0;
  if (chk_even != 3) {
    ans_mn = 1;
  } else {
    ans_mn = 3;
  }
  long long int cnt_adjL = 0;
  for (long long int i = 0; i < n; i++) {
    if (asNleaf[i]) ++cnt_adjL;
  }
  ans_mx = (n - 1) - leaf_cnt + cnt_adjL;
  cout << ans_mn << " " << ans_mx << '\n';
  return 0;
}

```