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;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
constexpr int SIZE = 1e5 + 1;
constexpr int PRSIZE = 1e4;
int primelist[PRSIZE], pcnt = 0;
bool isprime[SIZE];
int n;
void init() {
  for (int i = 2; i <= n; ++i) isprime[i] = true;
  for (int i = 2; i <= n; ++i) {
    if (isprime[i] == true) {
      primelist[pcnt++] = i;
      for (int j = 2; j * i <= n; ++j) isprime[i * j] = false;
    }
  }
}
inline int fpow(int a, int b) {
  if (b == 0) return 1;
  if (a == 0) return 0;
  int ans = 1, base = a;
  for (; b > 0; b /= 2) {
    if (b & 1) ans *= base;
    base = base * base;
  }
  return ans;
}
constexpr int BLOCKSIZE = 98;
int block[BLOCKSIZE], bcnt;
int main() {
  n = read();
  int sq = sqrt(double(n));
  init();
  int ans = 1;
  int LLCNT = n;
  for (int i = 0; i < pcnt; ++i) {
    int now = primelist[i];
    int answerA, answerB;
    if (primelist[i] <= sq) {
      cout << "B " << now << endl;
      cin >> answerB;
      LLCNT -= answerB;
      cout << "A " << now << endl;
      cin >> answerA;
      int factor = 1;
      if (answerA != 0) {
        LLCNT += answerA;
        int base = now;
        factor = base;
        for (int j = 2; base * now <= n; ++j) {
          base *= now;
          cout << "A " << base << endl;
          cin >> answerA;
          if (answerA == 1)
            factor = base;
          else
            break;
        }
        ans *= factor;
      }
    } else if (ans <= sqrt(n)) {
      cout << "B " << now << endl;
      block[bcnt++] = now;
      cin >> answerB;
      LLCNT -= answerB;
      if (bcnt == BLOCKSIZE || i == pcnt - 1) {
        cout << "A " << 1 << endl;
        int realcnt;
        cin >> realcnt;
        if (realcnt > LLCNT) {
          for (int ib = 0; ib < bcnt; ++ib) {
            cout << "A " << block[ib] << endl;
            cin >> answerA;
            if (answerA == 1) {
              ans *= block[ib];
              bcnt = 0;
              break;
            }
          }
        }
        bcnt = 0;
      }
    } else
      break;
  }
  cout << "C " << ans << endl;
  return 0;
}
