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 n, c1, c2 = -1, c3;
bitset<100001> b;
vector<long long> v;
bool b1;
int main() {
  cin >> n;
  c3 = n;
  for (long long i = 2; i <= n; i++) {
    if (!b[i]) {
      v.push_back(i);
      for (long long t = i * i; t <= n; t += i) b[t] = 1;
    }
  }
  for (int i = 0; i <= v.size() / 100; i++) {
    for (int t = i * 100; t < (i + 1) * 100 && t < v.size(); t++) {
      cout << "B " << v[t] << "\n";
      cout.flush();
      cin >> c1;
      c3 -= c1;
    }
    cout << "A " << 1 << "\n";
    cout.flush();
    cin >> c1;
    if (c1 >= c3 + 1) {
      c2 = i;
      break;
    }
  }
  if (c2 == -1) {
    cout << "C " << 1 << "\n";
    cout.flush();
    cin >> c1;
    return 0;
  }
  for (int t = c2 * 100; t < (c2 + 1) * 100 && t < v.size(); t++) {
    cout << "A " << v[t] << "\n";
    cout.flush();
    cin >> c1;
    if (c1 == 1) {
      c2 = t;
      c3 = v[t];
      break;
    }
  }
  for (int i = c2; i < v.size(); i++) {
    b1 = 1;
    while (b1 && c3 * v[i] <= n) {
      cout << "A " << c3 * v[i] << "\n";
      cout.flush();
      cin >> c1;
      if (c1 == 1)
        c3 *= v[i];
      else
        b1 = 0;
    }
  }
  cout << "C " << c3 << "\n";
  cout.flush();
  cin >> c1;
  return 0;
}
