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 int maxn = 1e5;
bool isprime[maxn + 3];
long long prime[maxn];
void sieve(int n) {
  isprime[0] = isprime[1] = false;
  for (int i = 2; i * i <= n; i++) {
    if (isprime[i]) {
      for (int j = i * i; j <= n; j += i) {
        isprime[j] = false;
      }
    }
  }
}
int main() {
  long long n;
  int i, j;
  memset(isprime, true, sizeof isprime);
  scanf("%lld", &n);
  sieve(n);
  int cnt = 0;
  for (i = 1; i <= n; i++) {
    if (isprime[i]) {
      prime[++cnt] = i;
    }
  }
  long long ans = 1;
  int root = (int)sqrt(cnt);
  bool vis[100001];
  memset(vis, false, sizeof vis);
  int curCnt = n;
  bool foundSmallest = false;
  for (i = 1; i <= cnt; i++) {
    if (i >= root && ans * prime[i - root + 1] > n) {
      break;
    }
    printf("B %lld\n", prime[i]);
    fflush(stdout);
    int numMul;
    scanf("%d", &numMul);
    int banyakMulTanpaX = 0;
    for (j = prime[i]; j <= n; j += prime[i]) {
      if (!vis[j]) {
        banyakMulTanpaX++;
        vis[j] = true;
        curCnt--;
      }
    }
    if (banyakMulTanpaX != numMul) {
      long long kali = prime[i];
      for (long long j = prime[i] * prime[i]; j <= n; j *= prime[i]) {
        printf("A %lld\n", j);
        fflush(stdout);
        int ada;
        scanf("%d", &ada);
        if (ada) {
          kali = j;
        } else {
          break;
        }
      }
      ans *= kali;
    }
    if ((i % root == 0 || i == cnt) && !foundSmallest) {
      printf("A 1\n");
      fflush(stdout);
      int sisaNum;
      scanf("%d", &sisaNum);
      if (sisaNum != curCnt) {
        for (j = i - root + 1; j <= i; j++) {
          printf("A %lld\n", prime[j]);
          fflush(stdout);
          int adaX;
          scanf("%d", &adaX);
          if (adaX) {
            long long kali = prime[j];
            for (long long k = prime[j] * prime[j]; k <= n; k *= prime[j]) {
              printf("A %lld\n", k);
              fflush(stdout);
              int ada;
              scanf("%d", &ada);
              if (ada) {
                kali = k;
              } else {
                break;
              }
            }
            ans *= kali;
            foundSmallest = true;
            break;
          }
        }
      }
    }
  }
  assert(ans <= n);
  printf("C %lld\n", ans);
  fflush(stdout);
  return 0;
};
