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>
enum Opcodes { A = 'A', B = 'B', C = 'C', MAX };
uint32_t askFor(Opcodes opcode, uint32_t num) {
  assert(opcode >= Opcodes::A && opcode < Opcodes::MAX);
  assert(num > 0);
  assert(opcode != Opcodes::B || num > 1);
  printf("%c %u\n", (char)opcode, num);
  fflush(stdout);
  uint32_t res = 0;
  if (opcode != Opcodes::C) {
    scanf("%u", &res);
  }
  return res;
}
uint32_t minSieve(bool *sieve, uint32_t last, uint32_t limit) {
  assert(sieve && last > 0);
  uint32_t res = 0;
  for (uint32_t i = last + 1; i < limit; i++) {
    if (sieve[i]) {
      res = i;
      break;
    }
  }
  return res;
}
uint32_t calc(uint32_t n) {
  constexpr size_t MAX_N = 100000;
  constexpr size_t SIEVE_SIZE = MAX_N + 1;
  static bool sieve[SIEVE_SIZE];
  for (size_t i = 0; i < SIEVE_SIZE; i++) {
    sieve[i] = true;
  }
  sieve[0] = sieve[1] = false;
  uint32_t min_sieve = 1;
  uint32_t x = 1;
  uint32_t limit = n;
  uint32_t remaining = n;
  const uint32_t group = (uint32_t)pow(n, 0.4);
  while (min_sieve > 0) {
    std::list<uint32_t> primes;
    for (uint32_t g = 0; g < group; g++) {
      min_sieve = minSieve(sieve, min_sieve, limit + 1);
      uint32_t p = min_sieve;
      if (p) {
        primes.push_back(p);
        sieve[p] = false;
        for (uint32_t i = p * p; i <= limit; i += p) {
          sieve[i] = false;
        }
        remaining -= askFor(Opcodes::B, p);
      } else {
        break;
      }
    }
    if (primes.size() > 0) {
      uint32_t real_count = askFor(Opcodes::A, 1);
      if (real_count != remaining) {
        remaining = real_count;
        for (auto p : primes) {
          uint32_t multiplier = 1;
          for (uint32_t power = 0; power < (uint32_t)(log(limit) / log(p));
               power++) {
            uint32_t num = pow(p, power + 1);
            if (askFor(Opcodes::A, num) > 0) {
              multiplier = num;
            } else {
              break;
            }
          }
          x *= multiplier;
          limit /= multiplier;
        }
      }
    }
  }
  return x;
}
int main() {
  uint32_t n, x = 1;
  scanf("%u", &n);
  if (n > 1) {
    x = calc(n);
  }
  askFor(Opcodes::C, x);
  return 0;
}
