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<vector<int>> conns;
void solve() {
  int n;
  cin >> n;
  conns.assign(n, vector<int>(0));
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    conns[a].push_back(b);
    conns[b].push_back(a);
  }
  auto leaf = [&](int i) { return conns[i].size() == 1; };
  vector<bool> ignored(n);
  int numig = 0;
  int anyleaf = -1;
  for (int i = 0; i < n; i++) {
    if (ignored[i]) continue;
    bool first = 0;
    for (int child : conns[i]) {
      if (leaf(child)) {
        anyleaf = child;
        if (!first) {
          first = 1;
        } else {
          numig += !ignored[child];
          ignored[child] = 1;
        }
      }
    }
  }
  assert(anyleaf >= 0);
  vector<bool> vis(n);
  queue<pair<int, int>> pq;
  pq.emplace(anyleaf, 0);
  bool hasUneven = 0;
  while (!pq.empty()) {
    int dist = pq.front().second;
    int pos = pq.front().first;
    pq.pop();
    if (vis[pos]) continue;
    vis[pos] = 1;
    if (leaf(pos) && dist % 2 == 1) hasUneven = 1;
    for (int p : conns[pos]) {
      pq.emplace(p, dist + 1);
    }
  }
  cout << (hasUneven ? 3 : 1) << " " << n - numig - 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int N = 1;
  for (int i = 0; i < N; i++) {
    solve();
  }
}
