#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > a(300000);
vector<bool> islist(300000, false);
vector<int> counoflist(300000, 0);
vector<int> len(300000, 0);
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    v1--;
    v2--;
    a[v1].push_back(v2);
    a[v2].push_back(v1);
  }
  for (int i = 0; i < n; i++) {
    if ((int)a[i].size() == 1) {
      islist[i] = true;
      counoflist[a[i][0]]++;
    }
  }
  int ma = n - 1;
  for (int i = 0; i < n; i++) {
    ma -= max(counoflist[i] - 1, 0);
  }
  int in = -1;
  for (int i = 0; i < n; i++) {
    if (islist[i]) in = i;
  }
  assert(in != -1);
  queue<int> q;
  q.push(in);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (auto &x : a[cur]) {
      if (len[x] == 0) {
        len[x] = len[cur] + 1;
        q.push(x);
      }
    }
  }
  int mi = 1;
  for (int i = 0; i < n; i++) {
    if (i == in) continue;
    if (len[i] % 2 == 1 && islist[i]) mi = 3;
  }
  cout << mi << " " << ma << "\n";
}

### Prompt

In cpp, your task is to solve 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<vector<int> > a(300000);
vector<bool> islist(300000, false);
vector<int> counoflist(300000, 0);
vector<int> len(300000, 0);
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    v1--;
    v2--;
    a[v1].push_back(v2);
    a[v2].push_back(v1);
  }
  for (int i = 0; i < n; i++) {
    if ((int)a[i].size() == 1) {
      islist[i] = true;
      counoflist[a[i][0]]++;
    }
  }
  int ma = n - 1;
  for (int i = 0; i < n; i++) {
    ma -= max(counoflist[i] - 1, 0);
  }
  int in = -1;
  for (int i = 0; i < n; i++) {
    if (islist[i]) in = i;
  }
  assert(in != -1);
  queue<int> q;
  q.push(in);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (auto &x : a[cur]) {
      if (len[x] == 0) {
        len[x] = len[cur] + 1;
        q.push(x);
      }
    }
  }
  int mi = 1;
  for (int i = 0; i < n; i++) {
    if (i == in) continue;
    if (len[i] % 2 == 1 && islist[i]) mi = 3;
  }
  cout << mi << " " << ma << "\n";
}

```