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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
int Ask(char op, int a) {
  cout << op << " " << a << "\n";
  cout.flush();
  if (op == 'C') {
    return 0;
  }
  int ans;
  cin >> ans;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  vector<bool> is_prime(n + 1, true);
  vector<int> p;
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; ++i) {
    if (is_prime[i]) {
      p.push_back(i);
      for (long long j = i * (long long)i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
  vector<int> small_p;
  vector<int> big_p;
  for (int i = 0; i < ((int)(p).size()); ++i) {
    if (p[i] * (long long)p[i] <= n) {
      small_p.push_back(p[i]);
    } else {
      big_p.push_back(p[i]);
    }
  }
  for (int i = 0; i < ((int)(small_p).size()); ++i) {
    Ask('B', small_p[i]);
  }
  int cur_total = Ask('A', 1);
  if (cur_total == ((int)(big_p).size()) + 2) {
    int res = 1;
    for (int i = 0; i < ((int)(small_p).size()); ++i) {
      int cur_pow = small_p[i];
      while (cur_pow <= n && Ask('A', cur_pow) > 0) {
        cur_pow *= small_p[i];
      }
      res *= (cur_pow / small_p[i]);
    }
    for (int i = 0; i < ((int)(big_p).size()); ++i) {
      if (Ask('A', big_p[i]) > 1) {
        res *= big_p[i];
      }
    }
    Ask('C', res);
    return 0;
  }
  vector<int> a = big_p;
  const int magic = 100;
  while (((int)(a).size()) > 0) {
    vector<int> del;
    while (((int)(a).size()) > 0 && ((int)(del).size()) < magic) {
      Ask('B', a.back());
      del.push_back(a.back());
      a.pop_back();
    }
    if (Ask('A', 1) > 1 + ((int)(a).size())) {
      for (int x : del) {
        if (Ask('A', x) > 0) {
          Ask('C', x);
          return 0;
        }
      }
    }
  }
  Ask('C', 1);
  return 0;
}
