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> v[100001], vis(100001), par(100001), w, dis(100001);
int cnt = 0;
void dfs(int i) {
  vis[i] = 1;
  int e = 0;
  for (auto x : v[i]) {
    if (!vis[x]) {
      e = 1;
      dis[x] = 1 + dis[i];
      cnt++;
      par[x] = i;
      dfs(x);
    }
  }
  if (e == 0) w.push_back(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (long long int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1);
  if (v[1].size() == 1) {
    w.push_back(1);
    par[1] = v[1][0];
  }
  set<int> s;
  int mi = 0;
  for (auto x : w) {
    s.insert(par[x]);
    if (dis[x] % 2) mi++;
  }
  if (mi == w.size() || mi == 0)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cnt -= (w.size() - s.size());
  cout << cnt;
  return 0;
}
