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::sync_with_stdio(0);
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 'C' << ' ' << n << endl;
    return 0;
  }
  vector<long long> a(n + 1, 1);
  vector<long long> b(n + 1, 1);
  vector<long long> sum(n + 1);
  vector<long long> p;
  for (int i = 2; i < n + 1; i++) {
    if (a[i]) {
      p.push_back(i);
      for (int j = i; j <= n; j += i) {
        sum[i] += a[j];
        a[j] = 0;
      }
    }
  }
  int cur = 1;
  int low = 0;
  int count = n;
  while (low < p.size()) {
    int mid = (low + p.size() + 1) / 2;
    if (low == 0) {
      mid = min(mid, 100);
    }
    for (int i = low; i < mid; i++) {
      cout << 'B' << ' ' << p[i] << endl;
      long long x;
      cin >> x;
      for (int j = p[i]; j <= n; j += p[i]) {
        b[j] = 0;
      }
      count -= x;
    }
    cout << 'A' << ' ' << 1 << endl;
    long long x;
    cin >> x;
    if (x > count) {
      if (low == 0) {
        mid = p.size();
      }
      for (int i = low; i < mid; i++) {
        long long pp = p[i];
        cout << 'A' << ' ' << pp << endl;
        long long x;
        cin >> x;
        int supposed = 0;
        for (int j = pp; j <= n; j += pp) {
          supposed += b[j];
        }
        while (x > supposed) {
          cur *= p[i];
          pp *= p[i];
          if (pp > n) {
            break;
          }
          cout << 'A' << ' ' << pp << endl;
          supposed = 0;
          for (int j = pp; j <= n; j += pp) {
            supposed += b[j];
          }
          cin >> x;
        }
      }
      break;
    } else {
      low = mid;
    }
  }
  cout << 'C' << ' ' << cur << endl;
  return 0;
  return 0;
}
