#include <bits/stdc++.h>
using namespace std;
long long powmod(long long a, long long b) {
  long long res = 1;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, u, v, root = 0, leaf = 0, order1 = 0, flag = 1;
vector<vector<int>> t;
int dfs(int v, int p) {
  bool is_order1 = 0, odd = 0, even = 0;
  int val = 1, mval = 1;
  for (auto b : t[v]) {
    if (b == p) continue;
    if ((int)(t[b]).size() == 1) {
      leaf++;
      is_order1 = 1;
      val = 1;
      odd = 1;
      continue;
    }
    val = dfs(b, v);
    if ((val & 1))
      odd = 1;
    else
      even = 1;
    mval = max(val, mval);
  }
  if (is_order1) order1++;
  if ((odd) && (even)) flag = 0;
  return mval + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  t.resize(n);
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    --u;
    --v;
    t[u].push_back(v);
    t[v].push_back(u);
    if ((int)(t[u]).size() > (int)(t[root]).size()) root = u;
    if ((int)(t[v]).size() > (int)(t[root]).size()) root = v;
  }
  dfs(root, -1);
  int fmx = n - 1 - leaf + order1;
  int fmn = flag ? 1 : 3;
  cout << fmn << ' ' << fmx << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long powmod(long long a, long long b) {
  long long res = 1;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, u, v, root = 0, leaf = 0, order1 = 0, flag = 1;
vector<vector<int>> t;
int dfs(int v, int p) {
  bool is_order1 = 0, odd = 0, even = 0;
  int val = 1, mval = 1;
  for (auto b : t[v]) {
    if (b == p) continue;
    if ((int)(t[b]).size() == 1) {
      leaf++;
      is_order1 = 1;
      val = 1;
      odd = 1;
      continue;
    }
    val = dfs(b, v);
    if ((val & 1))
      odd = 1;
    else
      even = 1;
    mval = max(val, mval);
  }
  if (is_order1) order1++;
  if ((odd) && (even)) flag = 0;
  return mval + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  t.resize(n);
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    --u;
    --v;
    t[u].push_back(v);
    t[v].push_back(u);
    if ((int)(t[u]).size() > (int)(t[root]).size()) root = u;
    if ((int)(t[v]).size() > (int)(t[root]).size()) root = v;
  }
  dfs(root, -1);
  int fmx = n - 1 - leaf + order1;
  int fmn = flag ? 1 : 3;
  cout << fmn << ' ' << fmx << '\n';
  return 0;
}

```