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;
long long int mod = 998244353;
long long int expo(long long int base, long long int exponent,
                   long long int mod) {
  long long int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
vector<long long int> g[100005];
long long int val[100005], dist[100005];
void dfs(long long int node, long long int par, long long int d) {
  dist[node] = d;
  if (g[node].size() == 1) val[par]++;
  for (auto x : g[node]) {
    if (x != par) dfs(x, node, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  while (tests--) {
    long long int i, j, u, v, n;
    cin >> n;
    for (long long int i = 1; i <= n - 1; i++) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    long long int root = -1;
    for (long long int i = 1; i <= n; i++) {
      if (g[i].size() > 1) {
        root = i;
        break;
      }
    }
    dfs(root, 0, 0);
    long long int mx = n - 1, mn;
    long long int odd = 0, even = 0;
    for (long long int i = 1; i <= n; i++) {
      if (g[i].size() > 1) continue;
      if (dist[i] % 2)
        odd++;
      else
        even++;
    }
    if (odd == 0 || even == 0)
      mn = 1;
    else
      mn = 3;
    for (long long int i = 1; i <= n; i++) {
      mx -= max(0LL, val[i] - 1);
    }
    cout << mn << " " << mx;
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
