#include <bits/stdc++.h>
using namespace std;
using llong = long long;
using uint = unsigned int;
const int maxN = 100002;
vector<int> nei[maxN];
int One(int v, int par = -1) {
  if (nei[v].size() == 1) return 0;
  set<int> s;
  for (int to : nei[v])
    if (to != par) {
      int curson = One(to, v);
      if (curson == -1) return -1;
      s.insert(curson);
    }
  if (s.size() > 1) return -1;
  return 1 ^ *s.begin();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    nei[u].push_back(v);
    nei[v].push_back(u);
  }
  int root = 1;
  for (int v = 1; v <= n; ++v)
    if (nei[v].size() > 1) root = v;
  if (One(root) != -1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  int mx = n - 1;
  for (int v = 1; v <= n; ++v) {
    int leafs = 0;
    for (int to : nei[v])
      if (nei[to].size() == 1) ++leafs;
    mx -= max(0, leafs - 1);
  }
  cout << mx;
}

### 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;
using llong = long long;
using uint = unsigned int;
const int maxN = 100002;
vector<int> nei[maxN];
int One(int v, int par = -1) {
  if (nei[v].size() == 1) return 0;
  set<int> s;
  for (int to : nei[v])
    if (to != par) {
      int curson = One(to, v);
      if (curson == -1) return -1;
      s.insert(curson);
    }
  if (s.size() > 1) return -1;
  return 1 ^ *s.begin();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    nei[u].push_back(v);
    nei[v].push_back(u);
  }
  int root = 1;
  for (int v = 1; v <= n; ++v)
    if (nei[v].size() > 1) root = v;
  if (One(root) != -1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  int mx = n - 1;
  for (int v = 1; v <= n; ++v) {
    int leafs = 0;
    for (int to : nei[v])
      if (nei[to].size() == 1) ++leafs;
    mx -= max(0, leafs - 1);
  }
  cout << mx;
}

```