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;
const int N = 1e5 + 5;
vector<int> G[N];
int n;
int d[N], v[N], fa[N], cnt[N], deep[N];
bool has_odd = false;
bool odd[N], even[N];
void dfs(int x, int dep, int f) {
  fa[x] = f;
  deep[x] = dep;
  bool is_leave = 1;
  for (auto y : G[x]) {
    if (y == f) continue;
    dfs(y, dep + 1, x);
    is_leave = 0;
  }
  if (is_leave) {
    cnt[fa[x]]++;
  }
}
void dp(int x, int f) {
  bool is_leave = 1;
  for (auto y : G[x]) {
    if (y == f) continue;
    is_leave = 0;
    dp(y, x);
    if (even[y]) odd[x] = 1;
    if (odd[y]) even[x] = 1;
  }
  if (is_leave) {
    odd[x] = 0;
    even[x] = 1;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int root = 1;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() > 1) root = i;
  }
  dfs(root, 0, 0);
  dp(root, 0);
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    if (odd[i] && even[i]) {
      ans = 3;
    }
  }
  cout << ans << " ";
  ans = n - 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) ans -= cnt[i] - 1;
  }
  cout << ans << endl;
  return 0;
}
