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, u, e, d[200005], mp[200005], x, y, m, k;
vector<int> v[200005], v1, v2;
void solve(int a, int l) {
  d[a] = d[l] + 1;
  if (v[a].size() == 1) {
    v1.push_back(a);
    mp[a] = 1;
  } else {
    v2.push_back(a);
  }
  for (int i = 0; i < v[a].size(); i++) {
    if (v[a][i] != l) {
      solve(v[a][i], a);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> e;
    v[u].push_back(e);
    v[e].push_back(u);
  }
  solve(1, 0);
  x = 1;
  for (int i = 1; i < v1.size(); i++) {
    if (d[v1[i - 1]] % 2 != d[v1[i]] % 2) {
      x = 3;
      break;
    }
  }
  for (int i = 0; i < v2.size(); i++) {
    y = v2[i];
    for (int j = 0; j < v[y].size(); j++) {
      if (mp[v[y][j]] == 1) {
        k++;
        break;
      }
    }
  }
  cout << x << " " << n - 1 - v1.size() + k;
}
