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.
#include <bits/stdc++.h>
const int C = 100005;
void supfactor(int* pr, int v) {
  int i, j;
  pr[1] = 1;
  for (i = 2; i * i < v; i++) {
    if (pr[i] == 0) {
      for (j = i; j < v; j += i) {
        if (pr[j] == 0) pr[j] = i;
      }
    }
  }
  for (; i < v; i++)
    if (pr[i] == 0) pr[i] = i;
}
int primes[C];
int main() {
  int n, i = 1, ij, slain = 0, cur = 0, con = 100, res = 1, x, primz = 1,
         lasti = 2, lastres, y;
  scanf("%d", &n);
  lastres = n;
  supfactor(primes, C);
  for (i = 2; i <= n;) {
    for (; i <= n; i++) {
      if (primes[i] != i) continue;
      printf("B %d\n", i), primz++;
      fflush(stdout);
      scanf("%d", &x);
      slain += x;
      if (primz % con == 0) break;
    }
    printf("A 1\n");
    fflush(stdout);
    scanf("%d", &y);
    if (y != lastres - slain) {
      for (ij = lasti; ij <= i; ij++) {
        if (primes[ij] != ij) continue;
        for (cur = ij; cur <= n; cur *= ij) {
          printf("A %d\n", cur);
          fflush(stdout);
          scanf("%d", &x);
          if (x == 0) break;
          res *= ij;
          if (cur == ij && ij > 1000) break;
        }
      }
    }
    lastres = y;
    slain = 0;
    i++;
    primz++;
    lasti = i;
  }
  printf("C %d\n", res);
  fflush(stdout);
  return 0;
}
