This is an interactive problem.

There is an unknown integer x (1≤ x≤ n). You want to find x.

At first, you have a set of integers \{1, 2, …, n\}. You can perform the following operations no more than 10000 times:

  * A a: find how many numbers are multiples of a in the current set. 
  * B a: find how many numbers are multiples of a in this set, and then delete all multiples of a, but x will never be deleted (even if it is a multiple of a). In this operation, a must be greater than 1. 
  * C a: it means that you know that x=a. This operation can be only performed once. 



Remember that in the operation of type B a>1 must hold.

Write a program, that will find the value of x.

Input

The first line contains one integer n (1≤ n≤ 10^5). The remaining parts of the input will be given throughout the interaction process.

Interaction

In each round, your program needs to print a line containing one uppercase letter A, B or C and an integer a (1≤ a≤ n for operations A and C, 2≤ a≤ n for operation B). This line desribes operation you make.

If your operation has type C your program should terminate immediately.

Else your program should read one line containing a single integer, which is the answer to your operation.

After outputting each line, don't forget to flush the output. To do it use:

  * fflush(stdout) in C/C++; 
  * System.out.flush() in Java; 
  * sys.stdout.flush() in Python; 
  * flush(output) in Pascal; 
  * See the documentation for other languages. 



It is guaranteed, that the number x is fixed and won't change during the interaction process.

Hacks:

To make a hack, use such input format:

The only line should contain two integers n, x (1 ≤ x ≤ n ≤ 10^5).

Example

Input


10

2

4

0

Output



B 4

A 2

A 8

C 4

Note

Note that to make the sample more clear, we added extra empty lines. You shouldn't print any extra empty lines during the interaction process.

In the first test n=10 and x=4.

Initially the set is: \{1,2,3,4,5,6,7,8,9,10\}.

In the first operation, you ask how many numbers are multiples of 4 and delete them. The answer is 2 because there are two numbers divisible by 4: \{4,8\}. 8 will be deleted but 4 won't, because the number x will never be deleted. Now the set is \{1,2,3,4,5,6,7,9,10\}.

In the second operation, you ask how many numbers are multiples of 2. The answer is 4 because there are four numbers divisible by 2: \{2,4,6,10\}.

In the third operation, you ask how many numbers are multiples of 8. The answer is 0 because there isn't any number divisible by 8 in the current set.

In the fourth operation, you know that x=4, which is the right answer.
import java.io.*;
import java.util.*;
 
public class Main extends PrintWriter {
    

    private void solve() {
        int n = sc.nextInt();
//        x = sc.nextInt(); removed = new boolean[n+1];
        boolean[] nums = new boolean[n+1];
        int ans = 1;
        for(int i = 2; i*i <= n; i++) {
            if(!nums[i]) {
                int ii = i;
                B(ii);
                while(ii <= n) {
                    int ret = A(ii);
                    if(ret == 0) break;
                    else ans *= i;
                    ii *= i;
                }
                for(int j = i; j <= n; j+= i) nums[j] = true;
            }
        }
        ArrayList<Integer> primes = new ArrayList<>();
        for(int i = 2; i <= n; i++) {
            if(!nums[i]) {
                primes.add(i);
            }            
        }
        if(ans > 1) {
            for(int x : primes) {
                if(B(x) > 1) {
                    ans *= x;
                }
            }
        }
        else {
            int m = primes.size();
            for(int i = 0, j = 0, cnt = 0; i <= m-1; i = j) {
                for(j = i; j <= (i+m-1)/2; j++) {
                    B(primes.get(j));
                    cnt++;
                }
                int ret = A(1);
                if(m-cnt < ret-1) {
                    for(j = i; j <= (i+m-1)/2; j++) {
                        if(A(primes.get(j)) > 0) {
                            ans *= primes.get(j);
                            C(ans);
                            return;
                        }
                    }
                }
            }
        }
        C(ans);
    }
    


    int A(int i) {
        println("A " + i);
        flush();
        return sc.nextInt();
    }
    
    int B(int i) {
        println("B " + i);
        flush();
        return sc.nextInt();
    }
    
    void C(int i) {
        println("C " + i);
        flush();
    }
    
//    boolean[] removed;
//    int x;
//    
//    
//    int op = 0;
//    int A(int i) {
//        op++;
//        int cnt = 0;
//        for(int j = i; j < removed.length; j+= i) if(!removed[j]) cnt++;
//        
//        return cnt;
//    }
//    
//    int B(int i) {
//        int cnt = A(i);
//        for(int j = i; j < removed.length; j+= i) if(j != x) removed[j] = true;
//        
//        return cnt;
//    }
//    
//    void C(int i) {
//        op++;
//        println("C " + i);
//        flush();
//    }
//    
    

//  Solution() throws FileNotFoundException { super(new File("output.txt")); }
//  InputReader sc = new InputReader(new FileInputStream("test_input.txt"));
    
  Main() { super(System.out); }
  InputReader sc = new InputReader(System.in);
  static class InputReader {
      InputReader(InputStream in) { this.in = in; } InputStream in;
      
      private byte[] buf = new byte[16384];
      private int    curChar;
      private int    numChars;
      
 
      public int read() {
          if (numChars == -1)
              throw new InputMismatchException();
          if (curChar >= numChars) {
              curChar = 0;
              try {
                  numChars = in.read(buf);
              } catch (IOException e) {
                  throw new InputMismatchException();
              }
              if (numChars <= 0)
                  return -1;
          }
          return buf[curChar++];
      }
 
      public String nextLine() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          StringBuilder res = new StringBuilder();
          do {
              res.appendCodePoint(c);
              c = read();
          } while (!isEndOfLine(c));
          return res.toString();
      }
 
      public String nextString() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          StringBuilder res = new StringBuilder();
          do {
              res.appendCodePoint(c);
              c = read();
          } while (!isSpaceChar(c));
          return res.toString();
      }
 
      public long nextLong() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          int sgn = 1;
          if (c == '-') {
              sgn = -1;
              c = read();
          }
          long 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 int nextInt() {
          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;
      }
 
      private boolean isSpaceChar(int c) {
          return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
      }
 
      private boolean isEndOfLine(int c) {
          return c == '\n' || c == '\r' || c == -1;
      }
  }
 
    public static void main(String[] $) {
        new Thread(null, new Runnable() {
            public void run() {
                long start = System.nanoTime();
                try {Main solution = new Main(); solution.solve(); solution.close();} 
                catch (Exception e) {e.printStackTrace(); System.exit(1);}
                System.err.println((System.nanoTime()-start)/1E9);
            }
        }, "1", 1 << 27).start();
 
    }
    

}