#include <bits/stdc++.h>
using namespace std;
void Fast_IO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
long long n;
vector<vector<long long> > adj;
vector<long long> sp;
long long oddpath;
void dfs(long long u, long long p, long long h) {
  if ((h % 2 == 1) && ((long long)int(adj[u].size()) == 1)) oddpath = 1;
  for (long long i = (long long)0;
       i <= (long long)(long long)int(adj[u].size()) - 1; i++)
    if (adj[u][i] != p) dfs(adj[u][i], u, h + 1);
}
int main() {
  Fast_IO();
  oddpath = 0;
  cin >> n;
  adj.resize(n);
  for (long long i = (long long)1; i <= (long long)n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long leaf = -1;
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) {
      leaf = i;
      break;
    }
  dfs(leaf, leaf, 0);
  long long maxv = n - 1;
  long long ln = 0;
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) ln++;
  maxv -= ln;
  sp.clear();
  for (long long i = (long long)0; i <= (long long)n - 1; i++) sp.push_back(0);
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) {
      sp[adj[i][0]] = 1;
    }
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if (sp[i]) maxv++;
  if (oddpath)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << maxv;
  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;
void Fast_IO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
long long n;
vector<vector<long long> > adj;
vector<long long> sp;
long long oddpath;
void dfs(long long u, long long p, long long h) {
  if ((h % 2 == 1) && ((long long)int(adj[u].size()) == 1)) oddpath = 1;
  for (long long i = (long long)0;
       i <= (long long)(long long)int(adj[u].size()) - 1; i++)
    if (adj[u][i] != p) dfs(adj[u][i], u, h + 1);
}
int main() {
  Fast_IO();
  oddpath = 0;
  cin >> n;
  adj.resize(n);
  for (long long i = (long long)1; i <= (long long)n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long leaf = -1;
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) {
      leaf = i;
      break;
    }
  dfs(leaf, leaf, 0);
  long long maxv = n - 1;
  long long ln = 0;
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) ln++;
  maxv -= ln;
  sp.clear();
  for (long long i = (long long)0; i <= (long long)n - 1; i++) sp.push_back(0);
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if ((long long)int(adj[i].size()) == 1) {
      sp[adj[i][0]] = 1;
    }
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    if (sp[i]) maxv++;
  if (oddpath)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << maxv;
  return 0;
}

```