#include <bits/stdc++.h>
using namespace std;
void dfs(int vertex, map<int, vector<int> >& mp1, vector<int>& dist) {
  for (int j = 1; j < mp1[vertex].size(); j++) {
    if (dist[mp1[vertex][j]] == -1) {
      dist[mp1[vertex][j]] = dist[vertex] + 1;
      dfs(mp1[vertex][j], mp1, dist);
    }
  }
}
int main() {
  int n;
  cin >> n;
  int x, y;
  map<int, vector<int> > mp1;
  set<int> inter, exter;
  for (int i = 1; i <= n; i++) mp1[i].push_back(-1);
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    mp1[x].push_back(y);
    mp1[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (mp1[i].size() > 2) {
      inter.insert(i);
    } else
      exter.insert(i);
  }
  int ind;
  set<int>::iterator itr;
  int inter_hav_ext = 0;
  for (itr = inter.begin(); itr != inter.end(); itr++) {
    int f = *itr;
    for (int j = 1; j < mp1[f].size(); j++)
      if (exter.find(mp1[f][j]) != exter.end()) {
        ind = f;
        inter_hav_ext++;
        mp1[f][0] = 0;
        break;
      }
  }
  long long max = 0, min = 1;
  vector<int> dist(n + 1, -1);
  dist[ind] = 0;
  dfs(ind, mp1, dist);
  for (int i = 1; i <= n; i++) {
    if (dist[i] % 2 == 1 && mp1[i][0] == 0) {
      min = 3;
      break;
    }
  }
  max = (n - 1 - exter.size()) + inter_hav_ext;
  cout << min << " " << max << 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;
void dfs(int vertex, map<int, vector<int> >& mp1, vector<int>& dist) {
  for (int j = 1; j < mp1[vertex].size(); j++) {
    if (dist[mp1[vertex][j]] == -1) {
      dist[mp1[vertex][j]] = dist[vertex] + 1;
      dfs(mp1[vertex][j], mp1, dist);
    }
  }
}
int main() {
  int n;
  cin >> n;
  int x, y;
  map<int, vector<int> > mp1;
  set<int> inter, exter;
  for (int i = 1; i <= n; i++) mp1[i].push_back(-1);
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    mp1[x].push_back(y);
    mp1[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (mp1[i].size() > 2) {
      inter.insert(i);
    } else
      exter.insert(i);
  }
  int ind;
  set<int>::iterator itr;
  int inter_hav_ext = 0;
  for (itr = inter.begin(); itr != inter.end(); itr++) {
    int f = *itr;
    for (int j = 1; j < mp1[f].size(); j++)
      if (exter.find(mp1[f][j]) != exter.end()) {
        ind = f;
        inter_hav_ext++;
        mp1[f][0] = 0;
        break;
      }
  }
  long long max = 0, min = 1;
  vector<int> dist(n + 1, -1);
  dist[ind] = 0;
  dfs(ind, mp1, dist);
  for (int i = 1; i <= n; i++) {
    if (dist[i] % 2 == 1 && mp1[i][0] == 0) {
      min = 3;
      break;
    }
  }
  max = (n - 1 - exter.size()) + inter_hav_ext;
  cout << min << " " << max << endl;
}

```