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<long long int> adj[100001];
long long int n, h[100001], ma, mi, least, o, e;
long long int dfs(long long int c, long long int p, long long int d) {
  int x, a, b = 0;
  for (int i = 0; i < adj[c].size(); i++) {
    x = adj[c][i];
    if (x != p) {
      a = dfs(x, c, !d);
      if (a == 1) b++;
      h[c] += a;
    }
  }
  if (!h[c]) {
    if (d)
      o = 1;
    else
      e = 1;
  }
  if (b == 1) b--;
  ma += b;
  if (b > 1) mi++;
  return h[c] + 1;
}
int main() {
  long long int x, y, i;
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  memset(h, 0, sizeof(h));
  ma = mi = 0;
  o = e = 0;
  for (i = 1; i <= n; i++) {
    if (adj[i].size() > 1) {
      x = dfs(i, 0, 0);
      break;
    }
  }
  if (o && e)
    least = 3;
  else
    least = 1;
  printf("%lld %lld", least, n - 1 - ma + mi);
  return 0;
}
