#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  vector<vector<int>> G;
  bool dfs(int v, int from, int d) {
    if (from != -1 && G[v].size() == 1) {
      return d % 2 == 0;
    }
    for (int ch : G[v]) {
      if (ch != from) {
        if (!dfs(ch, v, d + 1)) return false;
      }
    }
    return true;
  }
  void solve(int num, std::istream& in, std::ostream& out) {
    int n;
    in >> n;
    G.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      in >> a >> b;
      a--, b--;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    int lst = 0;
    set<int> pre;
    int v = -1;
    for (int i = 0; i < n; i++) {
      if (G[i].size() == 1) {
        pre.insert(G[i][0]);
        lst++;
        v = i;
      }
    }
    int rest = n - 1 - lst;
    int mx = rest + pre.size();
    int mn = 3;
    if (dfs(v, -1, 0)) {
      mn = 1;
    }
    out << mn << " " << mx;
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(0, in, out);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
class Solution {
 public:
  vector<vector<int>> G;
  bool dfs(int v, int from, int d) {
    if (from != -1 && G[v].size() == 1) {
      return d % 2 == 0;
    }
    for (int ch : G[v]) {
      if (ch != from) {
        if (!dfs(ch, v, d + 1)) return false;
      }
    }
    return true;
  }
  void solve(int num, std::istream& in, std::ostream& out) {
    int n;
    in >> n;
    G.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      in >> a >> b;
      a--, b--;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    int lst = 0;
    set<int> pre;
    int v = -1;
    for (int i = 0; i < n; i++) {
      if (G[i].size() == 1) {
        pre.insert(G[i][0]);
        lst++;
        v = i;
      }
    }
    int rest = n - 1 - lst;
    int mx = rest + pre.size();
    int mn = 3;
    if (dfs(v, -1, 0)) {
      mn = 1;
    }
    out << mn << " " << mx;
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(0, in, out);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}

```