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>
import java.io.*;
import java.util.*;

public class B {
  PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
  StringTokenizer tok;

  public void go() throws IOException {
    ntok();
    int n = ipar();
    Vertex[] vertices = new Vertex[n];
    for (int i = 0; i < n; i++) {
      vertices[i] = new Vertex(i + 1);
    }
    for (int i = 0; i < n - 1; i++) {
      ntok();
      int a = ipar() - 1;
      int b = ipar() - 1;
      vertices[a].others.add(vertices[b]);
      vertices[b].others.add(vertices[a]);
    }
    int max = 0;
    int min = 0;
    for (Vertex start : vertices) {
      if (start.others.size() != 1) {
        start.start = true;
        int[] ans = find(start);
        min = ans[2] == 0 ? 1 : 3;
        max = ans[1];
        break;
      }
    }
    out.printf("%d %d%n", min, max);

    out.flush();
    in.close();
  }

  public int[] find(Vertex curr) {
    if (curr.others.size() == 1) {
      // out.printf("%d : [0, 0, 0]%n", curr.i);
      return new int[] { 0, 0, 0 };
    }
    curr.visited = true;
    boolean one = false;
    boolean even = false;
    boolean odd = false;
    int sum = 0;
    int count = 0;
    int depth = 0;
    int oddChain = 0;
    for (Vertex other : curr.others) {
      if (other.visited) {
        continue;
      }
      int[] ret = find(other);
      sum += ret[1];
      depth = Math.max(ret[0] + 1, depth);
      if (ret[0] == 0) {
        if (!one) {
          one = true;
          count++;
        }
      } else {
        count++;
      }
      if (ret[0] % 2 == 0) {
        even = true;
      } else {
        odd = true;
      }
      oddChain |= ret[2];
    }
    int[] ans = new int[] { depth, sum + count, (even && odd ? 1 : 0) | oddChain };
    // out.printf("%d : %s%n", curr.i, Arrays.toString(ans));
    return ans;
  }

  private class Vertex {
    ArrayList<Vertex> others;
    int i;
    boolean visited;
    boolean start;

    public Vertex(int i) {
      others = new ArrayList<>();
      this.i = i;
    }
  }

  public void ntok() throws IOException {
    tok = new StringTokenizer(in.readLine());
  }

  public int ipar() {
    return Integer.parseInt(tok.nextToken());
  }

  public int[] iapar(int n) {
    int[] arr = new int[n];
    for (int i = 0; i < n; i++) {
      arr[i] = ipar();
    }
    return arr;
  }

  public long lpar() {
    return Long.parseLong(tok.nextToken());
  }

  public long[] lapar(int n) {
    long[] arr = new long[n];
    for (int i = 0; i < n; i++) {
      arr[i] = lpar();
    }
    return arr;
  }

  public double dpar() {
    return Double.parseDouble(tok.nextToken());
  }

  public String spar() {
    return tok.nextToken();
  }

  public static void main(String[] args) throws IOException {
    new B().go();
  }
}
