#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (b == 0) return (a);
  a %= b;
  return (gcd(b, a));
}
bool cmp(int a, int b) {
  if (a > b) {
    return (true);
  } else {
    return (false);
  }
}
int DFS(int node, vector<int>& height, vector<vector<int>>& graph) {
  int flag = 1;
  int flag2 = 1;
  for (long long int i = 0; i < graph[node].size(); i++) {
    if (height[graph[node][i]] == -1) {
      height[graph[node][i]] = height[node] + 1;
      flag = flag && DFS(graph[node][i], height, graph);
      flag2 = 0;
    }
  }
  if (flag2) {
    if (height[node] % 2 == 0) {
      return (1);
    } else {
      return (0);
    }
  }
  return (flag);
}
int main() {
  long long int n;
  cin >> n;
  vector<vector<int>> graph(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    int a;
    int b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  int root = 0;
  int count = 0;
  for (long long int i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      root = i;
      count++;
    }
  }
  vector<int> height(n + 1, -1);
  height[root] = 0;
  int flag = DFS(root, height, graph);
  int min = 3;
  if (flag) {
    min = 1;
  }
  int max = 0;
  for (long long int i = 1; i <= n; i++) {
    if (graph[i].size() > 1) {
      int flag = 0;
      for (long long int j = 0; j < graph[i].size(); j++) {
        if (graph[graph[i][j]].size() == 1) flag = 1;
      }
      max += flag;
    }
  }
  max += n - 1 - count;
  cout << min << " " << max;
}

### Prompt

Create a solution in cpp for 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;
int gcd(int a, int b) {
  if (b == 0) return (a);
  a %= b;
  return (gcd(b, a));
}
bool cmp(int a, int b) {
  if (a > b) {
    return (true);
  } else {
    return (false);
  }
}
int DFS(int node, vector<int>& height, vector<vector<int>>& graph) {
  int flag = 1;
  int flag2 = 1;
  for (long long int i = 0; i < graph[node].size(); i++) {
    if (height[graph[node][i]] == -1) {
      height[graph[node][i]] = height[node] + 1;
      flag = flag && DFS(graph[node][i], height, graph);
      flag2 = 0;
    }
  }
  if (flag2) {
    if (height[node] % 2 == 0) {
      return (1);
    } else {
      return (0);
    }
  }
  return (flag);
}
int main() {
  long long int n;
  cin >> n;
  vector<vector<int>> graph(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    int a;
    int b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  int root = 0;
  int count = 0;
  for (long long int i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      root = i;
      count++;
    }
  }
  vector<int> height(n + 1, -1);
  height[root] = 0;
  int flag = DFS(root, height, graph);
  int min = 3;
  if (flag) {
    min = 1;
  }
  int max = 0;
  for (long long int i = 1; i <= n; i++) {
    if (graph[i].size() > 1) {
      int flag = 0;
      for (long long int j = 0; j < graph[i].size(); j++) {
        if (graph[graph[i][j]].size() == 1) flag = 1;
      }
      max += flag;
    }
  }
  max += n - 1 - count;
  cout << min << " " << max;
}

```