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 main(void) {
  int n;
  cin >> n;
  vector<vector<int>> neighs(n, vector<int>());
  vector<int> degs(n, 0);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    neighs[a - 1].push_back(b - 1);
    neighs[b - 1].push_back(a - 1);
    degs[a - 1]++;
    degs[b - 1]++;
  }
  int root = -1;
  for (int i = 0; i < n; i++) {
    if (degs[i] > 1) {
      root = i;
      break;
    }
  }
  vector<bool> vis(n, false);
  vector<int> h(n, 0);
  vector<int> p(n, root);
  queue<int> toVisit;
  toVisit.push(root);
  while (!toVisit.empty()) {
    int v = toVisit.front();
    toVisit.pop();
    if (!vis[v]) {
      vis[v] = true;
      for (auto u : neighs[v]) {
        if (!vis[u]) {
          p[u] = v;
          h[u] = h[v] + 1;
          toVisit.push(u);
        }
      }
    }
  }
  bool hasOdd = false;
  bool hasEven = false;
  for (int i = 0; i < n; i++) {
    if (degs[i] == 1) {
      if (h[i] % 2 == 0) {
        hasEven = true;
      } else {
        hasOdd = true;
      }
    }
  }
  if (hasEven && hasOdd) {
    cout << "3 ";
  } else {
    cout << "1 ";
  }
  int maxCount = n - 1;
  vector<bool> checked(n, false);
  for (int i = 0; i < n; i++) {
    if (degs[i] == 1 && !checked[i]) {
      int par = p[i];
      for (auto u : neighs[par]) {
        if (u != p[par] && degs[u] == 1) {
          checked[u] = true;
          if (u != i) maxCount--;
        }
      }
    }
  }
  cout << maxCount << "\n";
}
