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;
int main() {
  ios_base::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  int n = 100000;
  vector<bool> P(n + 1);
  vector<long long> A, B;
  for (long long i = 2; i <= n; ++i)
    if (!P[i]) {
      if (i * i > n)
        B.push_back(i);
      else
        A.push_back(i);
      for (long long j = i * i; j <= n; j += i) P[j] = 1;
    }
  cin >> n;
  int res = 1;
  auto aksA = [&](int x) {
    cout << "A " << x << endl;
    cout.flush();
    int t;
    cin >> t;
    return t;
  };
  auto askB = [&](int x) {
    cout << "B " << x << endl;
    cout.flush();
    int t;
    cin >> t;
    return t;
  };
  for (int i = 0; i < ((int)(A).size()); ++i) {
    if (A[i] > n) continue;
    int t = askB(A[i]);
    t = aksA(A[i]);
    if (t == 1) {
      while (res * A[i] <= n && aksA(res * A[i])) res *= A[i];
    }
  }
  int a = 0, cur = aksA(1);
  for (int i = 0; i < 98; ++i) {
    int tmp = cur;
    for (int j = 0; j < 97; ++j, ++a) {
      if (B[a] > n) continue;
      int t = askB(B[a]);
      tmp -= t;
    }
    cur = aksA(1);
    if (tmp != cur) {
      for (int j = 0; j < 97; --j) {
        if (B[--a] > n) continue;
        int t = aksA(B[a]);
        if (t == 1) {
          res *= B[a];
          cout << "C " << res << endl;
          cout.flush();
          return 0;
        }
      }
    }
  }
  for (; a < ((int)(B).size()); ++a) {
    if (B[a] > n) continue;
    int t = askB(B[a]);
    t = aksA(B[a]);
    if (t == 1) {
      res *= B[a];
      break;
    }
  }
  cout << "C " << res << endl;
  cout.flush();
  return 0;
}
