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.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.List;
 import java.util.*;
 public class realfast implements Runnable 
  {
    private static final int INF = (int) 1e9;
    long in= 998244353;
    long fac[]= new long[3000];
    public  void solve() throws IOException 
    {
      
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
      boolean prime[] =new boolean[100001];

      Arrays.fill(prime,true);

      prime[1]=false;
      ArrayList<Integer> small=new ArrayList<>();
      ArrayList<Integer> big=new ArrayList<>();

      for(int i=2;i<=n;i++)
      {
        if(prime[i])
        {
            if(i>=400)
                big.add(i);
            else
                small.add(i);
        }
        for(int j=2*i;j<=n;j=j+i)
        {
            prime[j]= false;
        }

      }

      int a=1;

      for(int i =0;i<small.size();i++)
      {
        System.out.println("B "+small.get(i));
        System.out.flush();
        int a1 = in.nextInt();
        int low = small.get(i);
        for(int j=low;j<=n;j=j*low)
        {
            System.out.println("A "+j);
            System.out.flush();
            int b1 = in.nextInt();
            if(b1>0)
            {
                a=a*low;
            }
            else
                break;
        }


      }
      if(a>1){
      for(int i =0;i<big.size();i++)
      {
        System.out.println("B "+big.get(i));
        System.out.flush();
        int a1 = in.nextInt();

         

        if(a1>=2)
        {
           a=a*big.get(i);

        }        
       }

      }
      else
      {
        int root = big.size();
        root = (int)Math.pow(root,0.5);
        root++;
        for(int i =0;i<big.size();i++)
        {
               System.out.println("B "+big.get(i));
               System.out.flush();
               int a1 = in.nextInt();
               
               if(i==big.size()-1||i%root==root-1)
               {
                 System.out.println("A "+1);
                 System.out.flush();
                 int c1 = in.nextInt();
                 int kal =i/root;
                 if(c1!=big.size()-i)
                 {
                     for(int j =kal*root;j<kal*root+root&&j<big.size();j++)
                     {
                        System.out.println("A "+big.get(j));
                         System.out.flush();
                        int c2 = in.nextInt();
                         if(c2==1)
                         {
                           a=a*big.get(j);   
                           break;   
                         } 
                     }
                     break;
                 } 
                  
               }

        }

      }


      System.out.println("C "+a);

      
           
      

    }
   
    
    public int value (int seg[], int left , int right ,int index, int l, int r)
    {
            
            if(left>right)
            {
              return -100000000;
            }
            if(right<l||left>r)
                return -100000000;
            if(left>=l&&right<=r)
                return seg[index];
            int mid = left+(right-left)/2;
            int val = value(seg,left,mid,2*index+1,l,r);
            int val2 = value(seg,mid+1,right,2*index+2,l,r);
            return Math.max(val,val2);

    }
   
    public int gcd(int a , int b )
    {
      if(a<b)
      {
        int t =a;
        a=b;
        b=t;
      }
      if(a%b==0)
        return b ;
      return gcd(b,a%b);
    }
    public long pow(long n , long p,long m)
    {
         if(p==0)
            return 1;
        long val = pow(n,p/2,m);;
        val= (val*val)%m;
        if(p%2==0)
            return val;
        else
            return (val*n)%m;
    }
    
    
    ///////////////////////////////////////////////////////////////////////////////////////////////////////////////////
    public static void main(String[] args) {
        new Thread(null, new realfast(), "", 128 * (1L << 20)).start();
    }
 
    private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    private BufferedReader reader;
    private StringTokenizer tokenizer;
    private PrintWriter out;
 
    @Override
    public void run() {
        try {
            if (ONLINE_JUDGE || !new File("input.txt").exists()) {
                reader = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            } else {
                reader = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
            }
            solve();
        } catch (IOException e) {
            throw new RuntimeException(e);
        } finally {
            try {
                reader.close();
            } catch (IOException e) {
                // nothing
            }
            out.close();
        }
    }
 
    private String readString() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }
 
    @SuppressWarnings("unused")
    private int readInt() throws IOException {
        return Integer.parseInt(readString());
    }
 
    @SuppressWarnings("unused")
    private long readLong() throws IOException {
        return Long.parseLong(readString());
    }
 
    @SuppressWarnings("unused")
    private double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }
}
class edge implements Comparable<edge>{
    int val ;
    int color;
    
    edge(int u, int v)
    {
       this.val=u;
       this.color=v;
    }
    public int compareTo(edge e)
    {
        return this.val-e.val;
    }
}