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;
using namespace std::chrono;
int cm = 0;
void check(int curr, vector<vector<int>> &g, vector<bool> &v,
           vector<bool> &leaf) {
  v[curr] = true;
  bool ls = false;
  if (g[curr].size() == 1) {
    leaf[curr] = true;
  }
  for (auto c : g[curr]) {
    if (!v[c]) {
      check(c, g, v, leaf);
    }
    ls |= leaf[c];
  }
  if (ls) cm++;
}
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 0; i < n - 1; i++) {
    int f, t;
    cin >> f >> t;
    g[f - 1].push_back(t - 1);
    g[t - 1].push_back(f - 1);
  }
  int cv = 0;
  for (int i = 0; i < n; i++) {
    if (g[i].size() == 1) {
      cv = i;
      break;
    }
  }
  int minv;
  vector<bool> v(n, false);
  vector<int> dist(n, 0);
  queue<int> q;
  q.push(cv);
  v[cv] = true;
  dist[cv] = 0;
  while (!q.empty()) {
    int curr = q.front();
    q.pop();
    for (auto c : g[curr]) {
      if (!v[c]) {
        dist[c] = dist[curr] + 1;
        v[c] = true;
        q.push(c);
      }
    }
  }
  bool even = true;
  for (int i = 0; i < n; i++) {
    if (g[i].size() == 1 && dist[i] % 2 == 1) {
      even = false;
    }
  }
  if (even)
    minv = 1;
  else
    minv = 3;
  cm = 0;
  v.assign(n, false);
  vector<bool> leaf(n, false);
  leaf[cv] = true;
  check(cv, g, v, leaf);
  int lf = 0;
  for (int i = 0; i < n; i++)
    if (leaf[i]) lf++;
  cout << minv << " " << n - 1 - lf + cm << "\n";
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<milliseconds>(stop - start);
}
