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>
#include <bits/stdc++.h>
using namespace std;
int n, t[100009], no = 0, d = 0, odd = 0, pari = 0;
bool vis[100009];
vector<int> v[100009];
void dfs(int m) {
  vis[m] = true;
  t[m] = d;
  int h = 0;
  for (int i = 0; i < v[m].size(); i++) {
    if (!vis[v[m][i]]) {
      if (v[v[m][i]].size() == 1) h++;
      d++;
      dfs(v[m][i]);
      d--;
    }
  }
  h = max(0, h - 1);
  no += h;
  if (v[m].size() == 1) {
    pari = t[m] % 2;
  }
}
int main() {
  memset(vis, false, sizeof(vis));
  memset(t, 0, sizeof(t));
  cin >> n;
  int a, b, u = 1;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
    if (v[a].size() > 1) u = a;
    if (v[b].size() > 1) u = b;
  }
  dfs(u);
  for (int i = 1; i <= n; i++) {
    if (v[i].size() == 1 && (t[i] + pari) % 2 == 1) odd++;
  }
  (odd == 0) ? cout << "1 " : cout << "3 ";
  cout << n - 1 - no << endl;
}
