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;
int dfsRec(int n, map<int, vector<int>>& adj, map<int, bool>& visited,
           int depth, bool* cond) {
  int temp = 0;
  visited[n] = true;
  vector<int> values;
  bool ignoreOne = false;
  for (auto it = adj[n].begin(); it != adj[n].end(); it++) {
    if (!visited[*it]) {
      bool condTemp = false;
      int temp2 = dfsRec(*it, adj, visited, depth + 1, &condTemp) + 1;
      *cond = *cond || condTemp;
      if (temp2 == 1) {
        ignoreOne = ignoreOne || !values.empty();
        values.push_back(temp2);
      } else {
        temp += temp2;
      }
    }
  }
  if (depth != 1 && !ignoreOne) {
    for (auto it = values.begin(); it != values.end(); it++) {
      temp += *it;
    }
  } else if (depth != 1 || temp == 0) {
    temp += 1;
  }
  if (temp == 0) {
    *cond = (depth) % 2;
  }
  return temp;
}
int main() {
  int nodes;
  map<int, vector<int>> adj;
  cin >> nodes;
  for (int i = 0; i < nodes; i++) {
    int n1, n2;
    cin >> n1 >> n2;
    adj[n1].push_back(n2);
    adj[n2].push_back(n1);
  }
  map<int, bool> visited;
  bool cond = false;
  int i = 1;
  while (adj[i].size() != 1) i++;
  int x = dfsRec(i, adj, visited, 0, &cond);
  cout << (cond ? 3 : 1) << ' ' << (x > 2 ? x : 1) << endl;
}
