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;
vector<vector<int>> graph;
int got[2];
vector<bool> is_leaf;
int ans = 0;
void DFS(int node, int par, int depth) {
  bool has_leaf_son = false;
  for (auto vec : graph[node]) {
    if (vec == par) continue;
    is_leaf[node] = false;
    DFS(vec, node, depth + 1);
    if (!has_leaf_son || is_leaf[vec] == false) ans += 1;
    if (is_leaf[vec]) has_leaf_son = true;
  }
  if (is_leaf[node]) got[depth % 2] += 1;
}
int main() {
  int n;
  cin >> n;
  graph.resize(n);
  is_leaf.assign(n, true);
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  int root = 0;
  while (graph[root].size() == 1) ++root;
  DFS(root, -1, 0);
  int min_ans = (got[0] && got[1] ? 3 : 1);
  int max_ans = ans;
  cout << min_ans << " " << max_ans << endl;
  assert(min_ans <= max_ans);
  return 0;
}
