#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
ii solve(vector<vi> g) {
  int n = g.size();
  vector<int> depth(n, 0);
  vector<int> leaf(n, true);
  int mx = n - 1;
  function<void(int, int)> dfs = [&](int u, int p) {
    int cnt = 0;
    for (auto v : g[u]) {
      if (v == p) continue;
      depth[v] = depth[u] + 1;
      dfs(v, u);
      leaf[u] = false;
      if (leaf[v]) cnt++;
    }
    if (cnt) mx -= (cnt - 1);
  };
  int root = 0;
  for (int u = 0; u < n; ++u)
    if (g[u].size() > g[root].size()) root = u;
  ((void)0);
  dfs(root, -1);
  vector<int> parity(2, 0);
  for (int u = 0; u < n; ++u)
    if (leaf[u]) parity[depth[u] % 2]++;
  int mn = 1;
  if (parity[0] and parity[1]) mn = 3;
  return {mn, mx};
}
int main(void) {
  ios::sync_with_stdio(false), cin.tie(NULL);
  int n;
  cin >> n;
  vector<vi> g(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v, u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  auto [mn, mx] = solve(g);
  cout << mn << " " << mx << endl;
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
ii solve(vector<vi> g) {
  int n = g.size();
  vector<int> depth(n, 0);
  vector<int> leaf(n, true);
  int mx = n - 1;
  function<void(int, int)> dfs = [&](int u, int p) {
    int cnt = 0;
    for (auto v : g[u]) {
      if (v == p) continue;
      depth[v] = depth[u] + 1;
      dfs(v, u);
      leaf[u] = false;
      if (leaf[v]) cnt++;
    }
    if (cnt) mx -= (cnt - 1);
  };
  int root = 0;
  for (int u = 0; u < n; ++u)
    if (g[u].size() > g[root].size()) root = u;
  ((void)0);
  dfs(root, -1);
  vector<int> parity(2, 0);
  for (int u = 0; u < n; ++u)
    if (leaf[u]) parity[depth[u] % 2]++;
  int mn = 1;
  if (parity[0] and parity[1]) mn = 3;
  return {mn, mx};
}
int main(void) {
  ios::sync_with_stdio(false), cin.tie(NULL);
  int n;
  cin >> n;
  vector<vi> g(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v, u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  auto [mn, mx] = solve(g);
  cout << mn << " " << mx << endl;
  return 0;
}

```