#include <bits/stdc++.h>
using namespace std;
int n;
int du[100005] = {0};
int path[100005], cnt = 0, visited[100005] = {0};
vector<int> my_map[100005];
struct node {
  int xx, step;
};
void bfs(int now) {
  memset(path, -1, sizeof(path));
  path[now] = 0;
  queue<node> q;
  node into;
  into.xx = now;
  into.step = 0;
  q.push(into);
  while (!q.empty()) {
    node first = q.front();
    q.pop();
    int siz = my_map[first.xx].size();
    for (int i = 0; i < siz; ++i) {
      if (path[my_map[first.xx][i]] > 0) continue;
      into.xx = my_map[first.xx][i];
      into.step = first.step + 1;
      path[my_map[first.xx][i]] = into.step;
      q.push(into);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    ++du[a];
    ++du[b];
    my_map[a].push_back(b);
    my_map[b].push_back(a);
  }
  int s;
  for (int i = 1; i <= n; ++i)
    if (du[i] == 1) {
      s = i;
    }
  bfs(s);
  int flag = 0;
  for (int i = 1; i <= n; ++i) {
    int f = 0;
    int siz = my_map[i].size();
    for (int j = 0; j < siz; ++j)
      if (du[my_map[i][j]] == 1) ++f;
    if (f > 1) cnt += (f - 1);
    if (du[i] > 1 || i == s) continue;
    flag |= (path[i] & 1);
  }
  printf("%d %d\n", flag ? 3 : 1, n - 1 - cnt);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 n;
int du[100005] = {0};
int path[100005], cnt = 0, visited[100005] = {0};
vector<int> my_map[100005];
struct node {
  int xx, step;
};
void bfs(int now) {
  memset(path, -1, sizeof(path));
  path[now] = 0;
  queue<node> q;
  node into;
  into.xx = now;
  into.step = 0;
  q.push(into);
  while (!q.empty()) {
    node first = q.front();
    q.pop();
    int siz = my_map[first.xx].size();
    for (int i = 0; i < siz; ++i) {
      if (path[my_map[first.xx][i]] > 0) continue;
      into.xx = my_map[first.xx][i];
      into.step = first.step + 1;
      path[my_map[first.xx][i]] = into.step;
      q.push(into);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    ++du[a];
    ++du[b];
    my_map[a].push_back(b);
    my_map[b].push_back(a);
  }
  int s;
  for (int i = 1; i <= n; ++i)
    if (du[i] == 1) {
      s = i;
    }
  bfs(s);
  int flag = 0;
  for (int i = 1; i <= n; ++i) {
    int f = 0;
    int siz = my_map[i].size();
    for (int j = 0; j < siz; ++j)
      if (du[my_map[i][j]] == 1) ++f;
    if (f > 1) cnt += (f - 1);
    if (du[i] > 1 || i == s) continue;
    flag |= (path[i] & 1);
  }
  printf("%d %d\n", flag ? 3 : 1, n - 1 - cnt);
  return 0;
}

```