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;
long long inf = 987654321LL * 987654321LL;
bool used[100100];
vector<int> get_primes(int n) {
  vector<bool> is_prime(n + 1, true);
  for (int i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
  vector<int> result;
  for (int i = 2; i <= n; i++)
    if (is_prime[i]) result.push_back(i);
  return result;
}
int how_remove(int p, int n) {
  int ret = 0;
  for (int i = p; i <= n; i += p) {
    if (!used[i]) {
      ret++;
      used[i] = true;
    }
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  int batch_size = 100;
  auto primes = get_primes(n);
  int idx = 100;
  int ret;
  int past_exist = n;
  int remove_cnt = 0;
  vector<int> base;
  while (idx < primes.size()) {
    int st = idx;
    remove_cnt = 0;
    for (int ii = 0; ii < batch_size && idx < primes.size(); ii++, idx++) {
      printf("B %d\n", primes[idx]);
      fflush(stdout);
      scanf("%d", &ret);
      int rr = how_remove(primes[idx], n);
      remove_cnt += rr;
    }
    printf("A 1\n");
    fflush(stdout);
    int now_exist;
    scanf("%d", &now_exist);
    if (now_exist != past_exist - remove_cnt) {
      for (int i = st; i < idx; i++) {
        printf("A %d\n", primes[i]);
        fflush(stdout);
        scanf("%d", &ret);
        if (ret != 0) {
          base.push_back(primes[i]);
        }
      }
    }
    past_exist = now_exist;
  }
  for (int idx = 0; idx < min(100, (int)primes.size()); idx++) {
    printf("B %d\n", primes[idx]);
    fflush(stdout);
    scanf("%d", &ret);
    printf("A %d\n", primes[idx]);
    fflush(stdout);
    scanf("%d", &ret);
    if (ret != 0) base.push_back(primes[idx]);
  }
  map<int, int> degen;
  for (auto p : base) {
    degen[p] = 1;
    long long now = p;
    for (int i = 2; now * p <= n; i++) {
      now *= p;
      printf("A %lld\n", now);
      fflush(stdout);
      int exists = 0;
      scanf("%d", &exists);
      if (exists == 0)
        break;
      else
        degen[p] = i;
    }
  }
  long long ans = 1LL;
  for (auto p : base) {
    for (int i = 0; i < degen[p]; i++) {
      ans *= p;
    }
  }
  printf("C %lld\n", ans);
  fflush(stdout);
}
