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.
//stan hu tao
//join nct ridin by first year culture reps
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1406E
{
    static int MAX = 100000;
    static int SQRT = 100;
    public static void main(String hi[]) throws Exception
    {
        int[] div = new int[MAX+1];
        div[1] = 1;
        for(int d=2; d <= MAX; d++)
            for(int v=d; v <= MAX; v+=d)
                if(div[v] == 0)
                    div[v] = d;
        ArrayList<Integer> primes = new ArrayList<Integer>();
        for(int v=2; v <= MAX; v++)
            if(div[v] == v)
                primes.add(v);
        boolean[] removed = new boolean[MAX+1];
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());   MAX = N;
        if(N == 1)
        {
            out.println("C 1");
            return;
        }
        int active = N;
        ArrayList<Integer> factors = new ArrayList<Integer>();
        ArrayList<Integer> block = new ArrayList<Integer>();
        boolean firstFound = false;
        for(int q=0; q < primes.size(); q++)
        {
            int p = primes.get(q);
            if(p > N)
                break;
            block.add(p);
            int expected = 0;
            for(int v=p; v <= MAX; v+=p)
                if(!removed[v])
                    expected++;
            int cnt = askB(p, infile);
            for(int v=p; v <= MAX; v+=p)
                if(!removed[v])
                {
                    removed[v] = true;
                    active--;
                }
            if(expected != cnt)
                factors.add(p);
            if(!firstFound && block.size()%SQRT == 0)
            {
                cnt = askA(1, infile);
                if(cnt != active)
                {
                    for(int pp: block)
                        if(askA(pp, infile) == 1)
                        {
                            factors.add(pp);
                            break;
                        }
                    firstFound = true;
                }
                block = new ArrayList<Integer>();
            }
        }
        if(!firstFound)
        {
            for(int pp: block)
                if(askA(pp, infile) == 1)
                {
                    factors.add(pp);
                    break;
                }
            firstFound = true;
        }
        int res = 1;
        for(int p: factors)
        {
            long val = (long)p*p;
            while(val <= N)
            {
                int expected = 0;
                int act = (int)val;
                for(int v=act; v <= MAX; v+=act)
                    if(!removed[v])
                        expected++;
                if(expected == askA(act, infile))
                    break;
                val *= p;
            }
            res *= (int)(val/p);
        }
        out.println("C "+res);
    }
    public static int askA(int x, BufferedReader infile) throws Exception
    {
        out.println("A "+x);
        out.flush();
        return Integer.parseInt(infile.readLine());
    }
    public static int askB(int x, BufferedReader infile) throws Exception
    {
        out.println("B "+x);
        out.flush();
        return Integer.parseInt(infile.readLine());
    }
}