Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int main(void) {
  int n;
  std::cin >> n;
  const int mod = 100;
  bool isPrime[101010];
  std::fill_n(isPrime, 101010, true);
  isPrime[0] = isPrime[1] = false;
  int ans = 1;
  int cnt = 0;
  int prev = 2;
  int cntprimes = 0;
  for (int i = 2; i <= n; i++) {
    if (!isPrime[i]) continue;
    cntprimes++;
    for (int j = i + i; j <= n; j += i) {
      if (isPrime[j]) {
        isPrime[j] = false;
      }
    }
    std::cout << "B " << i << std::endl;
    int resp;
    std::cin >> resp;
    cnt += resp;
    if (cntprimes % mod == 0) {
      std::cout << "A " << 1 << std::endl;
      int resp;
      std::cin >> resp;
      if (resp != n - cnt) {
        for (int j = prev; j < i; j++) {
          if (!isPrime[j]) continue;
          long long k = j;
          while (k <= n) {
            std::cout << "B " << k << std::endl;
            int resp;
            std::cin >> resp;
            if (resp == 0) break;
            ans *= j;
            k *= j;
          }
        }
        cnt = n - resp;
      }
      prev = i;
    }
  }
  for (int j = prev; j <= n; j++) {
    if (!isPrime[j]) continue;
    long long k = j;
    while (k <= n) {
      std::cout << "B " << k << std::endl;
      int resp;
      std::cin >> resp;
      if (resp == 0) break;
      ans *= j;
      k *= j;
    }
  }
  std::cout << "C " << ans << std::endl;
  return 0;
}