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<int> g[200005];
int g1 = 0;
int o = 0;
void dfs(int p, int pr) {
  int i;
  int h = 0;
  for (i = 0; i < g[p].size(); i++) {
    int u = g[p][i];
    if (u != pr) {
      dfs(u, p);
      if (g[u].size() == 1) h++;
    }
  }
  if (h > 0) g1 += h - 1;
}
pair<bool, bool> dfs1(int p, int pr) {
  int i;
  pair<bool, bool> a, b;
  a.first = 0;
  a.second = 0;
  int h = 0;
  for (i = 0; i < g[p].size(); i++) {
    int u = g[p][i];
    if (u != pr) {
      h = 1;
      b = dfs1(u, p);
      a.first |= b.second;
      a.second |= b.first;
    }
  }
  if (a.first == 1 && a.second == 1) o = 1;
  if (h == 0) a.first = 1;
  return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i;
  for (i = 0; i < 2e5 + 5; i++) g[i].clear();
  int n;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int r = 1;
  for (i = 1; i <= n; i++)
    if (g[i].size() > 1) r = i;
  dfs(r, -1);
  pair<bool, bool> d = dfs1(1, -1);
  if (d.second == 1 && g[1].size() == 1) o = 1;
  if (o == 1)
    cout << 3;
  else
    cout << 1;
  cout << " " << n - 1 - g1;
  return 0;
}
