#include <bits/stdc++.h>
using namespace std;
template <class Ch, class Tr, class Container>
basic_ostream<Ch, Tr>& operator<<(basic_ostream<Ch, Tr>& os,
                                  const Container& x) {
  os << "{ ";
  bool f = 1;
  for (auto& y : x) {
    if (!f)
      os << " , ";
    else
      f = 0;
    os << y;
  }
  return os << "}";
}
template <class X, class Y>
ostream& operator<<(ostream& os, const pair<X, Y>& p) {
  return os << "[" << p.first << ", " << p.second << "]";
}
const int N = 1e5 + 5;
vector<int> adj[N];
bool odd[N], even[N];
bool dfs(int node, int par, int dep) {
  bool l = 1;
  for (int child : adj[node]) {
    if (child == par) continue;
    l = 0;
    if (dfs(child, node, dep + 1)) return 1;
    odd[node] |= odd[child];
    even[node] |= even[child];
  }
  if (l) {
    odd[node] = dep & 1;
    even[node] = !odd[node];
  }
  return odd[node] && even[node];
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int mx = n - 1, root = -1;
  vector<int> cnt(n + 1);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      cnt[adj[i].back()]++;
      mx -= cnt[adj[i].back()] > 1;
    } else
      root = i;
  }
  cout << (dfs(root, root, 0) ? 3 : 1) << ' ' << mx;
}

### Prompt

In CPP, your task is to solve 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;
template <class Ch, class Tr, class Container>
basic_ostream<Ch, Tr>& operator<<(basic_ostream<Ch, Tr>& os,
                                  const Container& x) {
  os << "{ ";
  bool f = 1;
  for (auto& y : x) {
    if (!f)
      os << " , ";
    else
      f = 0;
    os << y;
  }
  return os << "}";
}
template <class X, class Y>
ostream& operator<<(ostream& os, const pair<X, Y>& p) {
  return os << "[" << p.first << ", " << p.second << "]";
}
const int N = 1e5 + 5;
vector<int> adj[N];
bool odd[N], even[N];
bool dfs(int node, int par, int dep) {
  bool l = 1;
  for (int child : adj[node]) {
    if (child == par) continue;
    l = 0;
    if (dfs(child, node, dep + 1)) return 1;
    odd[node] |= odd[child];
    even[node] |= even[child];
  }
  if (l) {
    odd[node] = dep & 1;
    even[node] = !odd[node];
  }
  return odd[node] && even[node];
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int mx = n - 1, root = -1;
  vector<int> cnt(n + 1);
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      cnt[adj[i].back()]++;
      mx -= cnt[adj[i].back()] > 1;
    } else
      root = i;
  }
  cout << (dfs(root, root, 0) ? 3 : 1) << ' ' << mx;
}

```