#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int N = 5010;
std::vector<std::vector<long long>> v;
std::vector<long long> deg;
int solve() {
  long long n;
  cin >> n;
  deg.resize(n + 1, 0);
  v.resize(n + 1);
  std::vector<long long> dis(n + 1, 0);
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  long long k;
  for (long long i = 1; i <= n; i++)
    if (deg[i] > 1) {
      k = i;
      break;
    }
  queue<long long> q;
  long long d = 0;
  std::vector<bool> visit(n + 1, false);
  q.push(k);
  long long a = 0, b = 0;
  long long c = 0;
  std::vector<bool> mor(n + 1, false);
  while (!q.empty()) {
    long long temp = q.front();
    q.pop();
    visit[temp] = true;
    if (deg[temp] == 1) {
      d++;
      if (dis[temp] % 2)
        a++;
      else
        b++;
    }
    for (auto u : v[temp]) {
      if (deg[temp] > 1) {
        if (deg[u] == 1) mor[temp] = true;
      }
      if (!visit[u]) {
        q.push(u);
        dis[u] = dis[temp] + 1;
      }
    }
  }
  if (a == 0 && b != 0 || (a != 0 && b == 0))
    cout << 1 << " ";
  else
    cout << 3 << " ";
  for (long long i = 1; i <= n; i++)
    if (mor[i] == true) c++;
  cout << n - 1 - (d) + c << endl;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  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;
const int MOD = 1000000007;
const int N = 5010;
std::vector<std::vector<long long>> v;
std::vector<long long> deg;
int solve() {
  long long n;
  cin >> n;
  deg.resize(n + 1, 0);
  v.resize(n + 1);
  std::vector<long long> dis(n + 1, 0);
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  long long k;
  for (long long i = 1; i <= n; i++)
    if (deg[i] > 1) {
      k = i;
      break;
    }
  queue<long long> q;
  long long d = 0;
  std::vector<bool> visit(n + 1, false);
  q.push(k);
  long long a = 0, b = 0;
  long long c = 0;
  std::vector<bool> mor(n + 1, false);
  while (!q.empty()) {
    long long temp = q.front();
    q.pop();
    visit[temp] = true;
    if (deg[temp] == 1) {
      d++;
      if (dis[temp] % 2)
        a++;
      else
        b++;
    }
    for (auto u : v[temp]) {
      if (deg[temp] > 1) {
        if (deg[u] == 1) mor[temp] = true;
      }
      if (!visit[u]) {
        q.push(u);
        dis[u] = dis[temp] + 1;
      }
    }
  }
  if (a == 0 && b != 0 || (a != 0 && b == 0))
    cout << 1 << " ";
  else
    cout << 3 << " ";
  for (long long i = 1; i <= n; i++)
    if (mor[i] == true) c++;
  cout << n - 1 - (d) + c << endl;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```