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;
long long int n;
vector<vector<long long int> > adj;
long long int odd(0);
vector<long long int> dist;
long long int p_leaf(0), leaf(0), root(0);
unordered_map<long long int, long long int> deg;
void dfs(long long int node, long long int p) {
  long long int l = 0;
  for (int x : adj[node]) {
    if (x == root) ++l;
    if (x == p) continue;
    dist[x] = dist[node] + 1;
    dfs(x, node);
    if (deg[x] != 1) continue;
    ++l;
    if (dist[x] % 2 == 1) odd++;
  }
  leaf += l;
  if (l > 0) p_leaf++;
}
void mainSolve() {
  cin >> n;
  adj.resize(n + 1);
  dist.resize(n + 1, 0);
  for (int i = 1; i < n; i++) {
    long long int a;
    cin >> a;
    long long int b;
    cin >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  long long int f_min, f_max;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 1) {
      root = i;
      dfs(i, 0);
      f_min = (odd == 0 ? 1 : 3);
      f_max = (n - 1) - leaf + p_leaf;
      break;
    }
  }
  cout << f_min << " " << f_max << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    mainSolve();
  }
  return 0;
}
