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 P = 100;
int main() {
  int query = 0;
  int n;
  cin >> n;
  vector<bool> exist(n + 1, true);
  exist[0] = false;
  int sum = n;
  vector<bool> is_prime(n + 1, true);
  is_prime[0] = false;
  is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i * 2; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
  vector<int> prime;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) {
      prime.push_back(i);
    }
  }
  int cnt = prime.size();
  int x = 1;
  vector<int> pf;
  bool ok = false;
  for (int i = 0; i < (cnt + P - 1) / P; i++) {
    int L = P * i;
    int R = min(P * (i + 1), cnt);
    if (x * prime[L] > n) {
      break;
    }
    for (int j = L; j < R; j++) {
      if (x * prime[j] > n) {
        break;
      }
      query++;
      assert(query <= 10000);
      cout << "B " << prime[j] << endl;
      int res;
      cin >> res;
      int tmp = 0;
      for (int k = prime[j]; k <= n; k += prime[j]) {
        if (exist[k]) {
          exist[k] = false;
          sum--;
          tmp++;
        }
      }
      if (res != tmp) {
        x *= prime[j];
        pf.push_back(prime[j]);
      }
    }
    if (!ok) {
      query++;
      assert(query <= 10000);
      cout << "A 1" << endl;
      int res;
      cin >> res;
      if (res != sum) {
        for (int j = L; j < R; j++) {
          if (x * prime[j] > n) {
            break;
          }
          query++;
          assert(query <= 10000);
          cout << "A " << prime[j] << endl;
          int res;
          cin >> res;
          if (res == 1) {
            x *= prime[j];
            pf.push_back(prime[j]);
            break;
          }
        }
      }
      if (x > 1) {
        ok = true;
      }
    }
  }
  for (long long a : pf) {
    long long check = a * a;
    while (check <= n && a * x <= n) {
      query++;
      assert(query <= 10000);
      cout << "A " << check << endl;
      int res;
      cin >> res;
      if (res == 1) {
        x *= a;
      } else {
        break;
      }
      check *= a;
    }
  }
  query++;
  assert(query <= 10000);
  cout << "C " << x << endl;
}
