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;
const int maxn = 1e5 + 10;
int n, h[maxn], t, ans, E[2];
vector<int> v[maxn];
void dfs(int i, int j) {
  h[i] = h[j] + 1;
  int push_back = 0;
  if (v[i].size() == 1) E[h[i] & 1]++;
  for (auto A : v[i]) {
    if (A - j) {
      if (v[A].size() > 1)
        ans++;
      else
        push_back = 1;
      dfs(A, i);
    }
  }
  if (push_back) ans++;
}
void solve() {
  cin >> n;
  h[0] = 1;
  for (int y = 1; y <= n; y++) {
    h[y] = 0;
    v[y].clear();
  }
  for (int y = 1; y < n; y++) {
    int i, j;
    cin >> i >> j;
    v[i].push_back(j);
    v[j].push_back(i);
  }
  int root = 0;
  for (int y = 1; y <= n; y++)
    if (v[y].size() > 1) root = y;
  ans = E[1] = E[2] = 0;
  dfs(root, 0);
  if (E[0] == 0 || E[1] == 0)
    cout << 1;
  else
    cout << 3;
  cout << ' ' << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    solve();
  }
}
