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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Rustam Musin (t.me/musin_acm)
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    OutputWriter out = new OutputWriter(outputStream);
    BNaznachenieVesovRebram solver = new BNaznachenieVesovRebram();
    solver.solve(1, in, out);
    out.close();
  }

  static class BNaznachenieVesovRebram {
    List<Integer>[] g;
    boolean haveOdd;
    int maxCount;

    public void solve(int testNumber, InputReader in, OutputWriter out) {
      int n = in.readInt();
      g = new List[n];
      for (int i = 0; i < n; i++) {
        g[i] = new ArrayList<>();
      }
      for (int i = 0; i < n - 1; i++) {
        int u = in.readInt() - 1;
        int v = in.readInt() - 1;
        g[u].add(v);
        g[v].add(u);
      }
      int root = 0;
      while (g[root].size() == 1) root++;
      haveOdd = false;
      dfs1(root, -1, 0);
      int min = haveOdd ? 3 : 1;
      maxCount = n - 1;
      dfs2(root, -1);
      out.print(min, maxCount);
    }

    int dfs1(int v, int p, int depth) {
      if (g[v].size() == 1) {
        return depth;
      }
      int[] have = new int[2];
      for (int to : g[v]) if (to != p) have[dfs1(to, v, depth ^ 1)]++;
      if (have[0] != 0 && have[1] != 0) {
        haveOdd = true;
        return 0;
      }
      return have[0] != 0 ? 0 : 1;
    }

    void dfs2(int v, int p) {
      if (g[v].size() == 1) return;
      int c = 0;
      for (int to : g[v]) {
        if (g[to].size() == 1) c++;
        else if (to != p) dfs2(to, v);
      }
      maxCount -= Math.max(0, c - 1);
    }

  }

  static class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
      writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
      this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
      for (int i = 0; i < objects.length; i++) {
        if (i != 0) {
          writer.print(' ');
        }
        writer.print(objects[i]);
      }
    }

    public void close() {
      writer.close();
    }

  }

  static class InputReader {
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    private InputReader.SpaceCharFilter filter;

    public InputReader(InputStream stream) {
      this.stream = stream;
    }

    public int read() {
      if (numChars == -1) {
        throw new InputMismatchException();
      }
      if (curChar >= numChars) {
        curChar = 0;
        try {
          numChars = stream.read(buf);
        } catch (IOException e) {
          throw new InputMismatchException();
        }
        if (numChars <= 0) {
          return -1;
        }
      }
      return buf[curChar++];
    }

    public int readInt() {
      int c = read();
      while (isSpaceChar(c)) {
        c = read();
      }
      int sgn = 1;
      if (c == '-') {
        sgn = -1;
        c = read();
      }
      int res = 0;
      do {
        if (c < '0' || c > '9') {
          throw new InputMismatchException();
        }
        res *= 10;
        res += c - '0';
        c = read();
      } while (!isSpaceChar(c));
      return res * sgn;
    }

    public boolean isSpaceChar(int c) {
      if (filter != null) {
        return filter.isSpaceChar(c);
      }
      return isWhitespace(c);
    }

    public static boolean isWhitespace(int c) {
      return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public interface SpaceCharFilter {
      public boolean isSpaceChar(int ch);

    }

  }
}

