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;
int deg[100011];
int l[2];
vector<int> adj[100011];
int p[100011];
void dfs(int x, int parent, int parity) {
  p[x] = parent;
  if (adj[x].size() == 1) l[parity]++;
  if (parity)
    parity = 0;
  else
    parity = 1;
  for (auto c : adj[x]) {
    if (c != parent) dfs(c, x, parity);
  }
}
int main() {
  int n;
  cin >> n;
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  int x = 1;
  while (adj[x].size() == 1) x++;
  dfs(x, -1, 1);
  if (l[0] && l[1]) {
    cout << "3 ";
  } else
    cout << "1 ";
  int ans = n - 1;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) continue;
    int w = 0;
    for (int c : adj[i]) {
      if (c == p[i]) continue;
      if (adj[c].size() == 1) w++;
    }
    ans -= w ? w - 1 : 0;
  }
  cout << ans << endl;
  return 0;
}
