#include <bits/stdc++.h>
using namespace std;
using ULL = unsigned long long;
using UL = unsigned;
using LL = long long;
template <class Ty>
using passive_queue = priority_queue<Ty, vector<Ty>, greater<Ty>>;
struct Problem {
  void Solve() {
    UL N;
    cin >> N;
    vector<vector<UL>> E(N);
    for (UL i = 0; i < (N - 1); i++) {
      UL u, v;
      cin >> u >> v;
      u--;
      v--;
      E[u].push_back(v);
      E[v].push_back(u);
    }
    vector<UL> F(N);
    UL s;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() == 1) s = i;
    queue<UL> Q;
    F[s] = 1;
    Q.push(s);
    while (Q.size()) {
      UL p = Q.front();
      Q.pop();
      for (UL e : E[p]) {
        if (F[e]) continue;
        F[e] = 3 - F[p];
        Q.push(e);
      }
    }
    bool dual = false;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() == 1)
        if (F[i] == 2) dual = true;
    UL ans = (dual ? 3 : 1);
    cout << ans << " ";
    ans = N - 1;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() != 1) {
        UL n = 0;
        for (UL e : E[i])
          if (E[e].size() == 1) n++;
        n = max(n, 1u) - 1;
        ans -= n;
      }
    cout << ans << endl;
  }
  Problem();
};
int main() {
  unique_ptr<Problem> p(new Problem());
  p->Solve();
  return 0;
}
Problem::Problem() { cout << fixed << setprecision(10); }

### Prompt

Generate 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 ULL = unsigned long long;
using UL = unsigned;
using LL = long long;
template <class Ty>
using passive_queue = priority_queue<Ty, vector<Ty>, greater<Ty>>;
struct Problem {
  void Solve() {
    UL N;
    cin >> N;
    vector<vector<UL>> E(N);
    for (UL i = 0; i < (N - 1); i++) {
      UL u, v;
      cin >> u >> v;
      u--;
      v--;
      E[u].push_back(v);
      E[v].push_back(u);
    }
    vector<UL> F(N);
    UL s;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() == 1) s = i;
    queue<UL> Q;
    F[s] = 1;
    Q.push(s);
    while (Q.size()) {
      UL p = Q.front();
      Q.pop();
      for (UL e : E[p]) {
        if (F[e]) continue;
        F[e] = 3 - F[p];
        Q.push(e);
      }
    }
    bool dual = false;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() == 1)
        if (F[i] == 2) dual = true;
    UL ans = (dual ? 3 : 1);
    cout << ans << " ";
    ans = N - 1;
    for (UL i = 0; i < (N); i++)
      if (E[i].size() != 1) {
        UL n = 0;
        for (UL e : E[i])
          if (E[e].size() == 1) n++;
        n = max(n, 1u) - 1;
        ans -= n;
      }
    cout << ans << endl;
  }
  Problem();
};
int main() {
  unique_ptr<Problem> p(new Problem());
  p->Solve();
  return 0;
}
Problem::Problem() { cout << fixed << setprecision(10); }

```