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 pii = std::pair<int, int>;
using pll = std::pair<long long, long long>;
const int maxn = 1e5 + 100;
int deg[maxn];
std::vector<int> v[maxn];
int can1(int r, int p = -1) {
  if (v[r].size() == 1) return 0;
  int res = -1;
  for (int y : v[r]) {
    if (y == p) continue;
    int vv = can1(y, r);
    if (vv == -1) {
      return -1;
    }
    if (res == -1) {
      res = vv;
      continue;
    }
    if (res != vv) {
      return -1;
    }
  }
  return 1 - res;
}
int ans = 0;
void findmax(int r, int p = -1) {
  int fixed = 0;
  for (int y : v[r]) {
    if (y == p) continue;
    if (v[y].size() == 1)
      ++fixed;
    else
      findmax(y, r);
  }
  int total = v[r].size();
  if (p != -1) --total;
  if (fixed == 0) {
    ans += total;
  } else if (total != 0) {
    ans += 1 + total - fixed;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  std::cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    std::cin >> a >> b;
    deg[a]++;
    deg[b]++;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  int root = 0;
  for (int i = 1; i <= n && root == 0; i++)
    if (deg[i] > 1) root = i;
  findmax(root);
  std::cout << (can1(root) != -1 ? 1 : 3) << " " << ans << std::endl;
  return 0;
}
