#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
set<long long> s;
vector<long long> adj[100005];
long long maxx, minn = 1;
void dfs(long long node, long long pre, long long curr) {
  long long flag = 1;
  for (auto& i : adj[node]) {
    if (i != pre) {
      dfs(i, node, curr + 1);
      flag = 0;
    }
  }
  if (flag) {
    if (curr % 2) minn = 3;
    s.insert(pre);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, root, leaf = 0;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (long long i = 0; i < n; i++) {
    if (adj[i + 1].size() == 1) {
      root = i + 1;
      leaf++;
    }
  }
  dfs(root, 0ll, 0ll);
  for (auto& i : adj[root]) s.insert(i);
  maxx = n - 1 - leaf + s.size();
  cout << minn << " " << max(maxx, minn) << endl;
}

### Prompt

Generate a CPP solution 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;
const long long mod = 998244353;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
set<long long> s;
vector<long long> adj[100005];
long long maxx, minn = 1;
void dfs(long long node, long long pre, long long curr) {
  long long flag = 1;
  for (auto& i : adj[node]) {
    if (i != pre) {
      dfs(i, node, curr + 1);
      flag = 0;
    }
  }
  if (flag) {
    if (curr % 2) minn = 3;
    s.insert(pre);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, root, leaf = 0;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (long long i = 0; i < n; i++) {
    if (adj[i + 1].size() == 1) {
      root = i + 1;
      leaf++;
    }
  }
  dfs(root, 0ll, 0ll);
  for (auto& i : adj[root]) s.insert(i);
  maxx = n - 1 - leaf + s.size();
  cout << minn << " " << max(maxx, minn) << endl;
}

```