Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void in(T &x) {
  x = 0;
  char c = getchar();
  bool fu = 0;
  while (c < '0' || c > '9') {
    c = getchar();
    if (c == '-') fu = 1;
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  x = fu ? -x : x;
}
const double eps = 1e-6;
int n, m, t;
bool not_prime[200005];
int prime[200005], p_num;
int ksm(int x, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret *= x;
    p >>= 1;
    x *= x;
  }
  return ret;
}
bool check(int x, int p) {
  int t, pr = ksm(x, p);
  if (pr == 1) return 1;
  cout << "B " << pr << endl;
  fflush(stdout);
  cin >> t;
  cout << "A " << pr << endl;
  fflush(stdout);
  cin >> t;
  return t > 0;
}
void produce(int n) {
  not_prime[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!not_prime[i]) prime[p_num++] = i;
    for (int j = 0; j < p_num; ++j) {
      if (i * prime[j] > n) break;
      not_prime[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  cin >> n;
  produce(n);
  int base = 1;
  int start;
  for (int i = 0; i < p_num; ++i)
    if (prime[i] < 320) {
      int l = 0, r = (log(n) / log(prime[i]) + eps);
      ++r;
      while (r - l > 1) {
        int mid = l + r >> 1;
        if (check(prime[i], mid))
          l = mid;
        else
          r = mid;
      }
      base *= ksm(prime[i], l);
    } else if (base != 1) {
      if (base * prime[i] > n) break;
      int t;
      cout << "A " << base * prime[i] << endl;
      cin >> t;
      if (t > 0) {
        cout << "C " << base * prime[i] << endl;
        return 0;
      }
    } else {
      start = i;
      break;
    }
  if (base == 1) {
    int B = sqrt(p_num);
    for (int i = start; i < p_num; i += B) {
      int st = i, ed = min(i + B, p_num);
      int lst, now, t;
      cout << "A 1" << endl;
      fflush(stdout);
      cin >> lst;
      for (int j = st; j < ed; ++j) {
        cout << "B " << prime[j] << endl;
        fflush(stdout);
        cin >> t;
      }
      cout << "A 1" << endl;
      fflush(stdout);
      cin >> now;
      if (lst - now != ed - st) {
        for (int j = st; j < ed; ++j) {
          cout << "A " << prime[j] << endl;
          fflush(stdout);
          cin >> t;
          if (t > 0) {
            cout << "C " << prime[j] << endl;
            fflush(stdout);
            return 0;
          }
        }
      }
    }
  }
  cout << "C " << base << endl;
  fflush(stdout);
  return 0;
}