#include <bits/stdc++.h>
using namespace std;
void checkPossible(vector<vector<int>> &edges, int parent, int cur, int sum,
                   bool &possible) {
  if (parent == -1) {
    checkPossible(edges, cur, edges[cur][0], sum, possible);
  } else {
    sum++;
    if (edges[cur].size() == 1) {
      possible = (possible && ((sum % 2) == 0));
    } else {
      for (int j : edges[cur]) {
        if (j != parent) {
          checkPossible(edges, cur, j, sum, possible);
        }
      }
    }
  }
}
int main() {
  int N;
  cin >> N;
  vector<vector<int>> edges(N, vector<int>(0));
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int ansMin = 3;
  for (int i = 0; i < N; i++) {
    if (edges[i].size() == 1) {
      bool possible = true;
      checkPossible(edges, -1, i, 0, possible);
      if (possible) {
        ansMin = 1;
      }
      break;
    }
  }
  int ansMax = N - 1;
  for (int i = 0; i < N; i++) {
    int leafNeighbours = 0;
    for (int j = 0; j < edges[i].size(); j++) {
      if (edges[edges[i][j]].size() == 1) {
        leafNeighbours++;
      }
    }
    if (leafNeighbours > 1) {
      ansMax -= (leafNeighbours - 1);
    }
  }
  cout << ansMin << " " << ansMax << '\n';
}

### Prompt

Your task is to create 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 checkPossible(vector<vector<int>> &edges, int parent, int cur, int sum,
                   bool &possible) {
  if (parent == -1) {
    checkPossible(edges, cur, edges[cur][0], sum, possible);
  } else {
    sum++;
    if (edges[cur].size() == 1) {
      possible = (possible && ((sum % 2) == 0));
    } else {
      for (int j : edges[cur]) {
        if (j != parent) {
          checkPossible(edges, cur, j, sum, possible);
        }
      }
    }
  }
}
int main() {
  int N;
  cin >> N;
  vector<vector<int>> edges(N, vector<int>(0));
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int ansMin = 3;
  for (int i = 0; i < N; i++) {
    if (edges[i].size() == 1) {
      bool possible = true;
      checkPossible(edges, -1, i, 0, possible);
      if (possible) {
        ansMin = 1;
      }
      break;
    }
  }
  int ansMax = N - 1;
  for (int i = 0; i < N; i++) {
    int leafNeighbours = 0;
    for (int j = 0; j < edges[i].size(); j++) {
      if (edges[edges[i][j]].size() == 1) {
        leafNeighbours++;
      }
    }
    if (leafNeighbours > 1) {
      ansMax -= (leafNeighbours - 1);
    }
  }
  cout << ansMin << " " << ansMax << '\n';
}

```