#include <bits/stdc++.h>
using namespace std;
mt19937_64 mt(time(0));
const int MAXN = 1e5 + 10;
int deep[MAXN];
vector<int> edges[MAXN];
int ret = 0;
set<int> table;
void solve(int cur, int par) {
  deep[cur] = deep[par] + 1;
  if (edges[cur].size() == 1) table.insert(deep[cur] % 2);
  bool has_zero = false;
  for (int nxt : edges[cur]) {
    if (nxt == par) continue;
    solve(nxt, cur);
    if (edges[nxt].size() != 1)
      ret++;
    else
      has_zero = true;
  }
  if (has_zero) ret++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  int n, a, b;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int root = 1;
  while (edges[root].size() == 1) root++;
  solve(root, 0);
  if (table.size() != 2)
    cout << "1 ";
  else
    cout << "3 ";
  cout << ret << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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;
mt19937_64 mt(time(0));
const int MAXN = 1e5 + 10;
int deep[MAXN];
vector<int> edges[MAXN];
int ret = 0;
set<int> table;
void solve(int cur, int par) {
  deep[cur] = deep[par] + 1;
  if (edges[cur].size() == 1) table.insert(deep[cur] % 2);
  bool has_zero = false;
  for (int nxt : edges[cur]) {
    if (nxt == par) continue;
    solve(nxt, cur);
    if (edges[nxt].size() != 1)
      ret++;
    else
      has_zero = true;
  }
  if (has_zero) ret++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  int n, a, b;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int root = 1;
  while (edges[root].size() == 1) root++;
  solve(root, 0);
  if (table.size() != 2)
    cout << "1 ";
  else
    cout << "3 ";
  cout << ret << '\n';
  return 0;
}

```