#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
constexpr ll MOD = 1000 * 1000 * 1000 + 7;
constexpr int ALPHABET_SIZE = 26;
constexpr ll INF = 1e18;
pair<size_t, size_t> count_dp(int v, const vector<vector<int>>& g, int p = -1) {
  if (g[v].size() == 1) {
    return make_pair(1, 0);
  }
  pair<size_t, size_t> ans{0, 0};
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    auto cur = count_dp(u, g, v);
    ans.first += cur.second;
    ans.second += cur.first;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  size_t n;
  cin >> n;
  vector<vector<int>> g(n);
  for (size_t i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = 0;
  while (g[root].size() == 1) {
    ++root;
  }
  auto dp_res = count_dp(root, g);
  size_t min_ans = (dp_res.first == 0 || dp_res.second == 0) ? 1 : 3;
  unordered_map<int, size_t> par;
  for (size_t i = 0; i < n; ++i) {
    if (g[i].size() == 1) {
      ++par[g[i].front()];
    }
  }
  size_t max_ans = n - 1;
  for (auto [_, c] : par) {
    max_ans -= c - 1;
  }
  cout << min_ans << " " << max_ans << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 ull = unsigned long long;
constexpr ll MOD = 1000 * 1000 * 1000 + 7;
constexpr int ALPHABET_SIZE = 26;
constexpr ll INF = 1e18;
pair<size_t, size_t> count_dp(int v, const vector<vector<int>>& g, int p = -1) {
  if (g[v].size() == 1) {
    return make_pair(1, 0);
  }
  pair<size_t, size_t> ans{0, 0};
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    auto cur = count_dp(u, g, v);
    ans.first += cur.second;
    ans.second += cur.first;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  size_t n;
  cin >> n;
  vector<vector<int>> g(n);
  for (size_t i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = 0;
  while (g[root].size() == 1) {
    ++root;
  }
  auto dp_res = count_dp(root, g);
  size_t min_ans = (dp_res.first == 0 || dp_res.second == 0) ? 1 : 3;
  unordered_map<int, size_t> par;
  for (size_t i = 0; i < n; ++i) {
    if (g[i].size() == 1) {
      ++par[g[i].front()];
    }
  }
  size_t max_ans = n - 1;
  for (auto [_, c] : par) {
    max_ans -= c - 1;
  }
  cout << min_ans << " " << max_ans << "\n";
  return 0;
}

```