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>
int main() {
  using namespace std;
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<char> opts = vector<char>(n + 1, char(1));
  opts[0] = '\0';
  int optsLeft = n;
  int actionsTaken = 0;
  vector<int> primes;
  {
    vector<char> sieve = vector<char>(n + 1, char(1));
    for (int i = 2; i <= n; ++i) {
      if (sieve[i] == '\0') continue;
      primes.push_back(i);
      for (int j = i * 2; j <= n; j += i) sieve[j] = '\0';
    }
  }
  int currFactor = 1;
  int uncleanedActions = 0;
  int currPrime = 0;
  int firstUncheckedPrime = 0;
  int nPrimes = int(primes.size());
  constexpr int max_Actions = 10000;
  auto cleanup = [&]() {
    while (firstUncheckedPrime < nPrimes) {
      long long currQuery = 1LL * currFactor * primes[firstUncheckedPrime];
      if (currQuery > n) return true;
      ++actionsTaken;
      cout << "A " << currQuery << endl;
      int result;
      cin >> result;
      if (result > 0)
        currFactor *= primes[firstUncheckedPrime];
      else
        ++firstUncheckedPrime;
    }
    return currPrime == nPrimes;
  };
  auto B_sim = [&](int a) {
    int prevOptsLeft = optsLeft;
    for (int i = a; i <= n; i += a) {
      optsLeft -= opts[i];
      opts[i] = '\0';
    }
    return prevOptsLeft - optsLeft;
  };
  while (currPrime < nPrimes && 1LL * currFactor * primes[currPrime] <= n) {
    if (actionsTaken + 2 * uncleanedActions + 10 >= max_Actions) {
      assert(actionsTaken < max_Actions);
      ++actionsTaken;
      cout << "A 1\n";
      cout.flush();
      int result;
      cin >> result;
      if (result != optsLeft) {
        if (cleanup()) {
          ++actionsTaken;
          cout << "C " << currFactor << endl;
          return 0;
        } else {
          assert(currFactor <= n);
          assert(opts[currFactor] == '\0');
          opts[currFactor] = char(1);
          ++optsLeft;
        }
      }
      uncleanedActions = 0;
      firstUncheckedPrime = currPrime;
      continue;
    }
    int currQuery = currFactor * primes[currPrime++];
    ++actionsTaken;
    ++uncleanedActions;
    cout << "B " << currQuery << endl;
    int result;
    cin >> result;
    int expectedResult = B_sim(currQuery);
    if (expectedResult != result)
      ;
  }
  assert(cleanup());
  ++actionsTaken;
  cout << "C " << currFactor << endl;
  return 0;
}