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;
long long fix[200002], n, k, cur = 1, y[200002], C, q;
vector<int> v, g;
long long A(long long x) {
  cout << "A " << x << endl;
  cin >> x;
  return x;
}
long long B(long long x) {
  cout << "B " << x << endl;
  cin >> x;
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (long long i = 2; i <= 100000; i++)
    if (!fix[i])
      for (long long j = i * i; j <= 100000; j += i) fix[j] = 1;
  long long raod = 0;
  for (int i = 2; i <= n; i++)
    if (!fix[i]) v.push_back(i);
  long long t = 0, pir = 1;
  for (int i = 0; i < v.size(); i++) {
    if (v[i] <= 320) {
      k = B(v[i]);
      k = A(v[i]);
      if (k) {
        g.push_back(v[i]);
      }
    } else {
      k = B(v[i]);
      if (k >= 2) {
        g.push_back(v[i]);
        break;
      }
      t++;
      if (t % 100 == 0 || i == v.size() - 1) {
        k = A(1);
        if (k - 1 != (int)v.size() - i - 1 && pir) {
          pir = 0;
          int ind = i;
          for (int j = 1; j <= 100; j++) {
            k = A(v[ind]);
            if (k) {
              cout << "C " << v[ind] << endl;
              return 0;
            }
            ind--;
          }
        }
      }
    }
  }
  for (int i = 0; i < g.size(); i++) {
    long long x = g[i];
    while (1) {
      x *= g[i];
      if (x > n) break;
      k = A(x);
      if (!k) break;
    }
    cur *= x / g[i];
  }
  cout << "C " << cur << endl;
  return 0;
}
