#include <bits/stdc++.h>
using namespace std;
namespace IO {
template <class A, class B>
ostream &operator<<(ostream &out, pair<A, B> &a) {
  out << a.first << " " << a.second;
  return out;
}
template <class A, class B>
ostream &operator<<(ostream &out, vector<pair<A, B>> &a) {
  for (pair<A, B> x : a) out << x.first << " " << x.second << '\n';
}
template <class A>
ostream &operator<<(ostream &out, vector<A> &a) {
  for (A x : a) out << x << ' ';
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &x : a) in >> x;
  return in;
}
}  // namespace IO
using namespace IO;
const long long N = 1e5 + 3;
vector<long long> graph[N];
long long f = 1;
pair<long long, long long> dfs(long long v, long long p) {
  pair<long long, long long> res = {0, 0};
  if (graph[v].size() == 1) {
    res.first = 1;
  }
  for (long long x : graph[v])
    if (x != p) {
      pair<long long, long long> a = dfs(x, v);
      res.first += a.second;
      res.second += a.first;
    }
  if (res.first && res.second) f = 0;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = (0); i < (long long)(n - 1); ++i) {
    long long a, b;
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  long long min_res = 0;
  long long max_res = n - 1;
  for (long long i = (0); i < (long long)(n); ++i) {
    long long l = 0;
    for (long long v : graph[i]) {
      if (graph[v].size() == 1) ++l;
    }
    if (l > 1) max_res -= l - 1;
  }
  dfs(0, -1);
  if (f)
    cout << "1 " << max_res;
  else
    cout << "3 " << max_res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
}

### 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;
namespace IO {
template <class A, class B>
ostream &operator<<(ostream &out, pair<A, B> &a) {
  out << a.first << " " << a.second;
  return out;
}
template <class A, class B>
ostream &operator<<(ostream &out, vector<pair<A, B>> &a) {
  for (pair<A, B> x : a) out << x.first << " " << x.second << '\n';
}
template <class A>
ostream &operator<<(ostream &out, vector<A> &a) {
  for (A x : a) out << x << ' ';
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &x : a) in >> x;
  return in;
}
}  // namespace IO
using namespace IO;
const long long N = 1e5 + 3;
vector<long long> graph[N];
long long f = 1;
pair<long long, long long> dfs(long long v, long long p) {
  pair<long long, long long> res = {0, 0};
  if (graph[v].size() == 1) {
    res.first = 1;
  }
  for (long long x : graph[v])
    if (x != p) {
      pair<long long, long long> a = dfs(x, v);
      res.first += a.second;
      res.second += a.first;
    }
  if (res.first && res.second) f = 0;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = (0); i < (long long)(n - 1); ++i) {
    long long a, b;
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  long long min_res = 0;
  long long max_res = n - 1;
  for (long long i = (0); i < (long long)(n); ++i) {
    long long l = 0;
    for (long long v : graph[i]) {
      if (graph[v].size() == 1) ++l;
    }
    if (l > 1) max_res -= l - 1;
  }
  dfs(0, -1);
  if (f)
    cout << "1 " << max_res;
  else
    cout << "3 " << max_res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
}

```