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;
const int MAXN = 100000;
int p[MAXN + 5];
bool e[MAXN + 5], used[MAXN + 5];
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int cnt = 0;
  e[1] = true;
  for (int i = 2; i <= n; ++i) {
    if (!e[i]) {
      cnt++;
      p[cnt] = i;
      if (1ll * i * i <= 1ll * n) {
        for (int j = i * i; j <= n; j += i) {
          e[j] = true;
        }
      }
    }
  }
  bool flag = false;
  int s = sqrt(cnt), total = n, ans = 1, q;
  for (int i = 1; i <= cnt; ++i) {
    cout << "B " << p[i] << '\n';
    cout.flush();
    int c = 0;
    for (int j = p[i]; j <= n; j += p[i]) {
      if (!used[j]) {
        c++;
        total--;
        used[j] = 1;
      }
    }
    cin >> q;
    if (q != c) {
      for (long long j = p[i]; j <= 1ll * n; j *= p[i]) {
        cout << "A " << j << '\n';
        cout.flush();
        cin >> q;
        if (q > 0)
          ans *= p[i];
        else
          break;
      }
    }
    if ((i == cnt or i % s == 0) and !flag) {
      cout << "A " << 1 << '\n';
      cout.flush();
      cin >> q;
      if (q != total) {
        for (int j = i - s + 1; j <= i; ++j) {
          for (long long k = p[j]; k <= 1ll * n; k *= p[j]) {
            cout << "A " << k << '\n';
            cout.flush();
            cin >> q;
            if (q) {
              ans *= p[j];
              flag = true;
            } else
              break;
          }
          if (flag) break;
        }
      }
    }
  }
  cout << "C " << ans << '\n';
  cout.flush();
  return 0;
}
