#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5;
vector<int> g[MAXN + 1];
bool parity[MAXN + 1][2];
int mn = 1;
int dfs(int nod, int par) {
  int ans = (par != 0), sz = 0, cnt = 0;
  for (auto it : g[nod]) {
    if (it != par) {
      ans += dfs(it, nod);
      for (int p = 0; p < 2; p++) {
        parity[nod][p ^ 1] |= parity[it][p];
      }
      sz++;
      cnt += (g[it].size() == 1);
    }
  }
  if (sz == 0) {
    parity[nod][0] = 1;
  }
  if (parity[nod][0] && parity[nod][1]) {
    mn = 3;
  }
  if (cnt > 0) {
    ans -= (cnt - 1);
  }
  return ans;
}
int main() {
  int i, n;
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = 1;
  while ((int)g[root].size() == 1) {
    root++;
  }
  int mx = dfs(root, 0);
  cout << mn << " " << mx;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int MAXN = (int)1e5;
vector<int> g[MAXN + 1];
bool parity[MAXN + 1][2];
int mn = 1;
int dfs(int nod, int par) {
  int ans = (par != 0), sz = 0, cnt = 0;
  for (auto it : g[nod]) {
    if (it != par) {
      ans += dfs(it, nod);
      for (int p = 0; p < 2; p++) {
        parity[nod][p ^ 1] |= parity[it][p];
      }
      sz++;
      cnt += (g[it].size() == 1);
    }
  }
  if (sz == 0) {
    parity[nod][0] = 1;
  }
  if (parity[nod][0] && parity[nod][1]) {
    mn = 3;
  }
  if (cnt > 0) {
    ans -= (cnt - 1);
  }
  return ans;
}
int main() {
  int i, n;
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = 1;
  while ((int)g[root].size() == 1) {
    root++;
  }
  int mx = dfs(root, 0);
  cout << mn << " " << mx;
  return 0;
}

```