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;
vector<vector<long long> > v(100000 + 5);
long long ans;
long long f = 0;
long long dfs(long long u, long long p) {
  if (v[u].size() == 1) return 1;
  long long res = 0;
  for (auto it : v[u]) {
    if (it != p) {
      res += dfs(it, u);
    }
  }
  if (res) {
    ans -= (res - 1);
  }
  return 0;
}
pair<long long, long long> dfs2(long long u, long long p) {
  if (v[u].size() == 1) {
    return {1, 0};
  }
  pair<long long, long long> pr, res = {0, 0};
  for (auto it : v[u]) {
    if (it != p) {
      pr = dfs2(it, u);
      res.first |= pr.second;
      res.second |= pr.first;
    }
  }
  if (res.first + res.second == 2) {
    f = 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x, y, i, j;
  cin >> n;
  long long root;
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    if (v[i].size() != 1) {
      root = i;
      break;
    }
  }
  ans = n - 1;
  dfs(root, root);
  dfs2(root, root);
  cout << (f == 1 ? 3 : 1) << " " << ans;
  cerr << (1000 * clock()) / CLOCKS_PER_SEC << "ms";
  return 0;
}
