#include <bits/stdc++.h>
using namespace std;
void go(int parent, int node, int depth, int& odd_count, int& even_count,
        vector<vector<int>>& tree) {
  int count = 0;
  for (int i = 0; i < tree[node].size(); i++) {
    if (tree[node][i] != parent) {
      count++;
      go(node, tree[node][i], depth + 1, odd_count, even_count, tree);
    }
  }
  if (count == 0) {
    odd_count += depth % 2;
    even_count += !(depth % 2);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int>> tree(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    tree[u - 1].push_back(v - 1);
    tree[v - 1].push_back(u - 1);
  }
  int root = -1;
  for (int i = 0; i < tree.size(); i++) {
    if (tree[i].size() > 1) {
      root = i;
      break;
    }
  }
  int odd_count = 0, even_count = 0;
  go(-1, root, 0, odd_count, even_count, tree);
  int min_ans = 0, max_ans = n - 1;
  if (!odd_count || !even_count) {
    min_ans = 1;
  } else {
    min_ans = 3;
  }
  vector<int> parent_leaf_count(n, 0);
  for (int i = 0; i < n; i++) {
    if (tree[i].size() == 1) {
      parent_leaf_count[tree[i][0]]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (parent_leaf_count[i]) {
      max_ans -= (parent_leaf_count[i] - 1);
    }
  }
  printf("%d %d\n", min_ans, max_ans);
  return 0;
}

### 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 go(int parent, int node, int depth, int& odd_count, int& even_count,
        vector<vector<int>>& tree) {
  int count = 0;
  for (int i = 0; i < tree[node].size(); i++) {
    if (tree[node][i] != parent) {
      count++;
      go(node, tree[node][i], depth + 1, odd_count, even_count, tree);
    }
  }
  if (count == 0) {
    odd_count += depth % 2;
    even_count += !(depth % 2);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int>> tree(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    tree[u - 1].push_back(v - 1);
    tree[v - 1].push_back(u - 1);
  }
  int root = -1;
  for (int i = 0; i < tree.size(); i++) {
    if (tree[i].size() > 1) {
      root = i;
      break;
    }
  }
  int odd_count = 0, even_count = 0;
  go(-1, root, 0, odd_count, even_count, tree);
  int min_ans = 0, max_ans = n - 1;
  if (!odd_count || !even_count) {
    min_ans = 1;
  } else {
    min_ans = 3;
  }
  vector<int> parent_leaf_count(n, 0);
  for (int i = 0; i < n; i++) {
    if (tree[i].size() == 1) {
      parent_leaf_count[tree[i][0]]++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (parent_leaf_count[i]) {
      max_ans -= (parent_leaf_count[i] - 1);
    }
  }
  printf("%d %d\n", min_ans, max_ans);
  return 0;
}

```