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;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 998244353;
const long long N = 1e5 + 9;
const long long M = 1e6 + 9;
long long n, l, m, cnt, root;
vector<long long> G[N];
bool ll[N];
void dfs(long long u, long long f, long long dep) {
  bool ok = 0;
  for (auto v : G[u]) {
    if (ll[v]) ok = 1;
    if (v == f) continue;
    dfs(v, u, dep + 1);
  }
  if (ok && !ll[u]) m++;
  if (u == root || !ll[u]) return;
  if (~dep & 1) cnt++;
}
void solve() {
  cin >> n;
  for (long long i = 1, u, v; i <= n - 1; i++) {
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  for (long long u = 1; u <= n; u++) {
    if (G[u].size() == 1) {
      root = u;
      ll[u] = 1;
      l++;
    }
  }
  dfs(root, root, 1);
  if (cnt)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << n - 1 - l + m << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}
