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;
vector<int> pr;
void func() {
  vector<int> ipr(100005, 1);
  for (int i = 2; i < 100005; i++) {
    if (ipr[i]) {
      pr.push_back(i);
      for (int j = 2 * i; j < 100005; j += i) ipr[j] = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  func();
  int n;
  cin >> n;
  if (n == 1) {
    cout << "C 1" << endl;
    return 0;
  }
  int st = 0;
  vector<int> cr;
  int an = 1;
  for (int i = 0; pr[i] <= sqrt(n); i++) {
    int x1, x2;
    cout << "B " << pr[i] << endl;
    cin >> x1;
    cout << "B " << pr[i] << endl;
    cin >> x2;
    if (x2) cr.push_back(pr[i]), an *= pr[i];
    st = i + 1;
  }
  if (cr.size()) {
    for (int i = 0; i < cr.size(); i++) {
      while (an * cr[i] <= n) {
        cout << "A " << an * cr[i] << endl;
        int x;
        cin >> x;
        if (x)
          an *= cr[i];
        else
          break;
      }
    }
  }
  cout << "A 1" << endl;
  int rem;
  cin >> rem;
  while (1) {
    int cnt = 0;
    int x;
    while (cnt++ < 100 and pr[st] <= n) {
      cout << "B " << pr[st] << endl;
      cin >> x;
      if (x == 2) {
        cout << "C " << pr[st] * an << endl;
        return 0;
      }
      st++;
    }
    cnt--;
    rem -= cnt;
    cout << "A 1" << endl;
    cin >> x;
    if (x > rem) {
      for (int i = st - cnt; i < st; i++) {
        cout << "A " << pr[i] << endl;
        cin >> x;
        if (x) {
          cout << "C " << pr[i] << endl;
          return 0;
        }
      }
    }
    if (pr[st] > n) break;
  }
  cout << "C " << an << endl;
  return 0;
}
