#include <bits/stdc++.h>
using namespace std;
vector<int> mp[100009];
int d[100009], du[100009];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int k1, k2;
    cin >> k1 >> k2;
    mp[k1].push_back(k2);
    mp[k2].push_back(k1);
    du[k1]++;
    du[k2]++;
  }
  int beg1 = -1, beg2 = -1;
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) {
      beg1 = i;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      beg2 = i;
      break;
    }
  queue<int> q1, q2;
  q1.push(beg2);
  q2.push(1);
  while (!q1.empty()) {
    int now = q1.front();
    q1.pop();
    int c = q2.front();
    q2.pop();
    d[now] = c;
    for (int i = 0; i < mp[now].size(); i++) {
      if (d[mp[now][i]] == 0) {
        q1.push(mp[now][i]);
        q2.push(c + 1);
      }
    }
  }
  int bo = 0, cun = -1;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1) {
      if (cun == -1) {
        cun = d[i] % 2;
      } else {
        if ((d[i] + cun) % 2 == 1) bo = 1;
      }
    }
  }
  if (bo == 0)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  memset(d, 0, sizeof(d));
  int ans = n - 1;
  for (int i = 1; i <= n; i++) {
    int check = 0;
    for (int j = 0; j < mp[i].size(); j++) {
      if (du[mp[i][j]] == 1) {
        if (check == 0)
          check = 1;
        else
          ans--;
      }
    }
  }
  cout << ans << 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> mp[100009];
int d[100009], du[100009];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int k1, k2;
    cin >> k1 >> k2;
    mp[k1].push_back(k2);
    mp[k2].push_back(k1);
    du[k1]++;
    du[k2]++;
  }
  int beg1 = -1, beg2 = -1;
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) {
      beg1 = i;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      beg2 = i;
      break;
    }
  queue<int> q1, q2;
  q1.push(beg2);
  q2.push(1);
  while (!q1.empty()) {
    int now = q1.front();
    q1.pop();
    int c = q2.front();
    q2.pop();
    d[now] = c;
    for (int i = 0; i < mp[now].size(); i++) {
      if (d[mp[now][i]] == 0) {
        q1.push(mp[now][i]);
        q2.push(c + 1);
      }
    }
  }
  int bo = 0, cun = -1;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1) {
      if (cun == -1) {
        cun = d[i] % 2;
      } else {
        if ((d[i] + cun) % 2 == 1) bo = 1;
      }
    }
  }
  if (bo == 0)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  memset(d, 0, sizeof(d));
  int ans = n - 1;
  for (int i = 1; i <= n; i++) {
    int check = 0;
    for (int j = 0; j < mp[i].size(); j++) {
      if (du[mp[i][j]] == 1) {
        if (check == 0)
          check = 1;
        else
          ans--;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```