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;
class TaskE {
  int64_t n;
  vector<int64_t> primes;
  vector<bool> isPrime;
  int64_t Ask(int64_t candidate, bool deleteMode) {
    int64_t res;
    if (deleteMode) {
      cout << 'B' << ' ' << candidate << endl;
      cin >> res;
    } else {
      cout << 'A' << ' ' << candidate << endl;
      cin >> res;
    }
    return res;
  }

 public:
  TaskE(int64_t n) {
    this->n = ++n;
    isPrime.assign(n, true);
    isPrime[0] = isPrime[1] = false;
    for (int64_t i = 2; i < n; i++) {
      if (isPrime[i]) {
        for (int64_t j = i * i; j < n; j += i) {
          isPrime[j] = false;
        }
      }
    }
    for (int64_t i = 0; i < n; i++) {
      if (isPrime[i]) primes.emplace_back(i);
    }
  }
  void solve() {
    int64_t res = 1;
    vector<int64_t> bigPrimes;
    for (int64_t prime : primes) {
      int64_t primeWithPower = prime;
      if (prime * prime < n) {
        Ask(primeWithPower, true);
        while (primeWithPower < n and Ask(primeWithPower, false) > 0) {
          res *= prime;
          primeWithPower *= prime;
        }
      } else {
        bigPrimes.emplace_back(prime);
      }
    }
    if (res != 1) {
      for (int64_t prime : bigPrimes) {
        if (res * prime < n and Ask(res * prime, true) > 0) {
          res *= prime;
          break;
        }
      }
    } else {
      int64_t rem = Ask(1, false);
      while (!bigPrimes.empty()) {
        int64_t sz = bigPrimes.size();
        int64_t hf = sz / 2;
        for (int64_t cnd = hf; cnd < sz; cnd++) Ask(bigPrimes[cnd], true);
        int64_t cur = Ask(1, false);
        if (rem - cur != sz - hf) {
          for (int64_t cnd = hf; cnd < sz; cnd++) {
            if (Ask(bigPrimes[cnd], false) > 0) {
              res *= bigPrimes[cnd];
              break;
            }
          }
          break;
        }
        rem = cur;
        bigPrimes.resize(hf);
      }
    }
    cout << 'C' << ' ' << res << endl;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t n;
  cin >> n;
  TaskE E(n);
  E.solve();
}
