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>
using namespace std;
const double eps = 1e-10;
const double PI = 3.1415926535897932;
const int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};
const int maxn = 1e5 + 5;
int n;
int prime[maxn], pn = 0;
bool is_prime[maxn];
int ans = 1;
int main() {
  scanf("%d", &n);
  memset(is_prime, true, sizeof(is_prime));
  is_prime[0] = is_prime[1] = true;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) prime[++pn] = i;
    for (int j = 1; j <= pn && i * prime[j] <= n; j++) {
      is_prime[i * prime[j]] = false;
      if (i % prime[j] == 0) break;
    }
  }
  int st = 0;
  for (int i = 1; i <= pn && prime[i] <= n / prime[i]; i++) {
    int res;
    printf("B %d\n", prime[i]);
    fflush(stdout);
    scanf("%d", &res);
    long long t = prime[i];
    while (t <= n) {
      printf("A %d\n", t);
      fflush(stdout);
      scanf("%d", &res);
      if (res == 0) {
        ans *= t / prime[i];
        break;
      }
      if (t * prime[i] > n) {
        ans *= t;
        break;
      }
      t *= prime[i];
    }
    st = i;
  }
  int sum;
  printf("A 1\n");
  fflush(stdout);
  scanf("%d", &sum);
  int block = 100;
  int cnt = 0;
  int first = st + 1;
  for (int i = st + 1; i <= pn; i++) {
    int res;
    printf("B %d\n", prime[i]);
    fflush(stdout);
    scanf("%d", &res);
    if (res == 2) {
      ans *= prime[i];
      break;
    }
    cnt++;
    if (cnt == block || i == pn) {
      printf("A 1\n");
      fflush(stdout);
      scanf("%d", &res);
      if (res != sum - (i - st)) {
        for (int j = first; j <= i; j++) {
          printf("A %d\n", prime[j]);
          fflush(stdout);
          scanf("%d", &res);
          if (res == 1) {
            ans *= prime[j];
            break;
          }
        }
        break;
      }
      first = i + 1;
      cnt = 0;
    }
  }
  printf("C %d\n", ans);
  fflush(stdout);
  return 0;
}
