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 N = 1e5 + 5;
bool prime[N];
int n;
vector<int> a;
int main() {
  int n;
  cin >> n;
  int res = 1;
  for (int i = 2; i <= n; i++)
    if (!prime[i]) {
      for (int j = 2; i * j <= n; j++) prime[i * j] = true;
      if (1LL * i * i <= n) {
        int j = 1;
        int tmp = 1;
        while (i * tmp <= n) {
          int cnt;
          cout << "B " << i * tmp << endl;
          cin >> cnt;
          cout << "A " << i * tmp << endl;
          cin >> cnt;
          if (cnt == 0) break;
          j++;
          tmp *= i;
        }
        res *= tmp;
      } else
        a.push_back(i);
    }
  if (res == 1) {
    int pos = 0;
    int m = a.size();
    while (m > 0 && a[pos] * res <= n) {
      int len = (m + 1) / 2;
      for (int i = 0; i < len; i++) {
        int tmp;
        cout << "B " << a[pos + i] << endl;
        cin >> tmp;
      }
      int cnt = 0;
      cout << "A " << 1 << endl;
      cin >> cnt;
      if (cnt - 1 != m - len) {
        bool flag = false;
        for (int i = 0; i < len; i++) {
          int cnt;
          cout << "A " << a[pos + i] << endl;
          cin >> cnt;
          if (cnt == 1) {
            res *= a[pos + i];
            flag = true;
            break;
          }
        }
        if (flag) break;
      } else {
        m -= len;
        pos += len;
      }
    }
  } else {
    for (int i = 0; i < (int)a.size(); i++) {
      int cnt;
      cout << "A " << a[i] << endl;
      cin >> cnt;
      if (cnt == 2) res *= a[i];
    }
  }
  cout << "C " << res << endl;
  return 0;
}
