Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int CHUNK = 50;
int n;
vector<int> primes;
void sieve() {
  int composite[n + 1];
  memset(composite, 0, sizeof composite);
  for (int i = 2; i <= n; i++) {
    if (composite[i]) continue;
    for (long long j = 1ll * i * i; j <= n; j += i) composite[j] = 1;
    primes.push_back(i);
  }
}
int querya() {
  int res;
  cout << "A 1" << endl;
  cin >> res;
  return res;
}
int queryb(int num) {
  int res;
  cout << "B " << num << endl;
  cin >> res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  sieve();
  int ct = 0;
  int tot = n;
  vector<int> chunks;
  for (int i = 0; i < primes.size() + 1; i++) {
    if (i != 0 && (i % CHUNK == 0 || i == primes.size())) {
      int actual = querya();
      if (actual > tot) chunks.push_back(ct);
      tot = actual;
      ct++;
    }
    if (i == primes.size()) break;
    tot -= queryb(primes[i]);
  }
  assert(chunks.size() <= 3);
  int ans = 1;
  for (int c : chunks) {
    for (int i = c * CHUNK; i < min((c + 1) * CHUNK, (int)primes.size()); i++) {
      int w = queryb(primes[i]);
      if (w != 0) {
        long long tmp = 1ll * primes[i] * primes[i];
        ans *= primes[i];
        while (tmp <= n && queryb(tmp)) {
          ans *= primes[i];
          tmp *= primes[i];
        }
      }
    }
  }
  cout << "C " << ans << endl;
}