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 n;
const int maxN = 1e5 + 3;
vector<int> g[maxN];
vector<int> dfs(int v, int par, bool& good, int& sub) {
  bool even = false, odd = false;
  vector<int> rr;
  int to_sub = 0;
  for (int ne : g[v]) {
    if (ne == par) {
      continue;
    }
    vector<int> r = dfs(ne, v, good, sub);
    for (int e : r) {
      if (e % 2) {
        odd = true;
      } else {
        even = true;
      }
      if (e == 0) {
        to_sub++;
      }
      rr.push_back(e + 1);
    }
  }
  if (to_sub) {
    sub += to_sub - 1;
  }
  if (odd && even) {
    good = true;
  }
  if (rr.empty()) {
    rr.push_back(0);
  }
  return rr;
}
int main() {
  cin >> n;
  int start;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
    if (g[u].size() > 1) {
      start = u;
    }
    if (g[v].size() > 1) {
      start = v;
    }
  }
  bool has_odd = false;
  int sub = 0;
  dfs(start, -1, has_odd, sub);
  int cnt = n - 1 - sub;
  if (has_odd) {
    cout << 3 << ' ' << max(3, cnt) << '\n';
  } else {
    cout << 1 << ' ' << cnt << '\n';
  }
  return 0;
}
