#include <bits/stdc++.h>
using namespace std;
vector<long long> v[200000];
long long vis[200000], z = 0;
vector<long long> k;
void dfs(long long x, long long y) {
  long long t = 0;
  if (v[x].size() == 1) k.push_back(y);
  vis[x] = 1;
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] == 1 && v[1].size() == 1) t++;
    if (vis[v[x][i]] == 0) {
      if (v[v[x][i]].size() == 1) t++;
      dfs(v[x][i], y + 1);
    }
  }
  if (t > 0) z++;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long n, x, y, m = 0;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (long long i = 1; i < n + 1; i++) vis[i] = 0;
  dfs(1, 0);
  for (long long i = 1; i < k.size(); i++) {
    if (k[i - 1] % 2 != k[i] % 2) {
      m++;
      break;
    }
  }
  if (m > 0)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << (n - 1) - k.size() + z;
}

### 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;
vector<long long> v[200000];
long long vis[200000], z = 0;
vector<long long> k;
void dfs(long long x, long long y) {
  long long t = 0;
  if (v[x].size() == 1) k.push_back(y);
  vis[x] = 1;
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] == 1 && v[1].size() == 1) t++;
    if (vis[v[x][i]] == 0) {
      if (v[v[x][i]].size() == 1) t++;
      dfs(v[x][i], y + 1);
    }
  }
  if (t > 0) z++;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long n, x, y, m = 0;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (long long i = 1; i < n + 1; i++) vis[i] = 0;
  dfs(1, 0);
  for (long long i = 1; i < k.size(); i++) {
    if (k[i - 1] % 2 != k[i] % 2) {
      m++;
      break;
    }
  }
  if (m > 0)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << (n - 1) - k.size() + z;
}

```