#include <bits/stdc++.h>
using namespace std;
vector<int> g[100100];
int n;
int res[100100];
int path_odd[100100];
int path_even[100100];
int degree[100100];
vector<int> v[100100];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
    degree[x]++;
    degree[y]++;
  }
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (g[i].size() == 1) {
      q.push(i);
      v[i].push_back(0);
      path_even[i] = 1;
    }
  }
  int mn = 1, mx = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    degree[x]--;
    res[x] = 0;
    sort(v[x].begin(), v[x].end());
    res[x] += v[x][0];
    for (int i = 1; i < v[x].size(); i++)
      if (v[x][i] != 1) res[x] += v[x][i];
    mx = max(res[x], mx);
    for (int i = 0; i < g[x].size(); i++) {
      int c = g[x][i];
      if (degree[c] == 0) {
        if (path_even[c]) path_odd[x] = 1;
        if (path_odd[c]) path_even[x] = 1;
      } else {
        degree[c]--;
        v[c].push_back(res[x] + 1);
        if (degree[c] == 1) q.push(c);
      }
    }
    if (path_odd[x] && path_even[x]) mn = 3;
  }
  cout << mn << " " << mx << endl;
  return 0;
}

### 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;
vector<int> g[100100];
int n;
int res[100100];
int path_odd[100100];
int path_even[100100];
int degree[100100];
vector<int> v[100100];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
    degree[x]++;
    degree[y]++;
  }
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (g[i].size() == 1) {
      q.push(i);
      v[i].push_back(0);
      path_even[i] = 1;
    }
  }
  int mn = 1, mx = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    degree[x]--;
    res[x] = 0;
    sort(v[x].begin(), v[x].end());
    res[x] += v[x][0];
    for (int i = 1; i < v[x].size(); i++)
      if (v[x][i] != 1) res[x] += v[x][i];
    mx = max(res[x], mx);
    for (int i = 0; i < g[x].size(); i++) {
      int c = g[x][i];
      if (degree[c] == 0) {
        if (path_even[c]) path_odd[x] = 1;
        if (path_odd[c]) path_even[x] = 1;
      } else {
        degree[c]--;
        v[c].push_back(res[x] + 1);
        if (degree[c] == 1) q.push(c);
      }
    }
    if (path_odd[x] && path_even[x]) mn = 3;
  }
  cout << mn << " " << mx << endl;
  return 0;
}

```