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 n;
bool found[300004];
vector<long long> graph[300004];
long long deg[300004];
bool odd;
bool even;
long long depth[300004];
bool vis[300004];
void dfs(long long cur, long long par, long long steps, long long st) {
  if (steps > 2) return;
  vis[cur] = true;
  for (int i = 0; i < graph[cur].size(); i++) {
    long long v = graph[cur][i];
    if (v == par) continue;
    dfs(v, cur, steps + 1, st);
  }
  return;
}
void init(long long cur, long long par) {
  depth[cur] = depth[par] + 1;
  for (int i = 0; i < graph[cur].size(); i++) {
    long long v = graph[cur][i];
    if (v == par) continue;
    init(v, cur);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  long long root = 1;
  for (int i = 1; i < n + 1; i++) {
    if (deg[i] > 1) {
      root = i;
      break;
    }
  }
  init(root, root);
  bool can = true;
  for (int i = 1; i < n + 1; i++) {
    if (deg[i] == 1) {
      if (depth[i] % 2 == 1)
        odd = true;
      else
        even = true;
    }
  }
  can = !(odd && even);
  if (can)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  long long leaves = 0, g = 0;
  for (int i = 1; i < n + 1; i++) {
    if (deg[i] == 1) {
      leaves++;
      if (!vis[i]) {
        g++;
        dfs(i, i, 0, i);
      }
    }
  }
  long long ans = (n - 1) - leaves + g;
  cout << ans << endl;
  return 0;
}
