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 <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int vis[100009], mark[100009], n, X, q;
int B(int x) {
  if (x > n) return 0;
  q++;
  printf("B %d", x);
  cout << endl;
  int res;
  scanf("%d", &res);
  return res;
}
int A(int x) {
  if (x > n) return 0;
  q++;
  printf("A %d", x);
  cout << endl;
  int res;
  scanf("%d", &res);
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) mark[i] = 1;
  vector<int> pr, v;
  for (int i = 2; i <= n; i++) {
    if (vis[i]) continue;
    pr.push_back(i);
    if (i >= 317) v.push_back(i);
    for (int j = i; j <= n; j += i) vis[j] = 1;
  }
  int ans = 1;
  for (int i = 0; i < int(pr.size()) and pr[i] < 317; i++) {
    B(pr[i]);
    for (int j = 0; j < 20 and B(ans * pr[i]); j++) ans *= pr[i];
  }
  swap(v, pr);
  if (ans > 1) {
    for (__typeof((pr).begin()) it = (pr).begin(); it != (pr).end(); it++)
      if (B(*it) == 2) {
        ans *= (*it);
        break;
      }
  } else {
    int tmp = A(1);
    for (int i = 0; i < int(pr.size()); i += 100) {
      for (int j = 0; j < 100 and i + j < int(pr.size()); j++)
        B(pr[i + j]), tmp--;
      if (A(1) != tmp) {
        for (int j = 0; j < 100 and i + j < int(pr.size()); j++)
          if (B(pr[i + j])) {
            ans *= pr[i + j];
            break;
          }
        break;
      }
    }
  }
  printf("C %d", ans);
  cout << endl;
  cerr << "q"
       << " = " << q << endl;
  ;
  return 0;
}