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;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& x) {
  return out << "(" << x.first << "," << x.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << " ] " << endl;
}
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args) {
  const string name = names.substr(0, names.find(','));
  cerr << name << " : " << arg1 << " | ";
  __trace__(names.substr(1 + (int)name.size()), args...);
}
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int maxn = 1e5 + 5;
vector<int> adj[maxn];
bool odd[maxn], even[maxn];
int mx = 0;
bool fl = 0;
int dfs(int u, int p) {
  int num = 0, ans = 0;
  for (int v : adj[u]) {
    if (v != p) {
      if ((int)adj[v].size() == 1) {
        num++;
        odd[u] = 1;
      } else {
        ans += dfs(v, u);
        if (odd[v] and odd[u]) {
          fl = 1;
        }
        if (even[u] and even[v]) {
          fl = 1;
        }
        odd[u] |= even[v];
        even[u] |= odd[v];
        ans++;
      }
    }
  }
  if (num)
    return ans + 1;
  else
    return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(12);
  cout << fixed;
  int n;
  cin >> n;
  for (int _ = 0; _ < n - 1; _++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int mx;
  for (int i = 1; i <= n; i++) {
    if ((int)adj[i].size() != 1) {
      mx = dfs(i, -1);
      break;
    }
  }
  int mn = (fl ? 3 : 1);
  cout << mn << " " << mx << '\n';
  time_taken();
  return 0;
}
