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;
using ll = int64_t;
int query(char c, int x) {
  cout << c << ' ' << x << endl;
  if (c == 'C') exit(0);
  int ans;
  cin >> ans;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> primes;
  for (int i = 2; i <= n; ++i) {
    bool good = true;
    for (int x : primes) {
      if (i % x == 0) {
        good = 0;
        break;
      }
    }
    if (good) primes.push_back(i);
  }
  int ans = 1;
  for (int x : primes) {
    if (x * x > n) break;
    query('B', x);
    for (ll deg = x; deg <= n; deg *= x)
      if (query('A', deg))
        ans *= x;
      else
        break;
  }
  reverse((primes).begin(), (primes).end());
  while (!primes.empty() && primes.back() * primes.back() <= n)
    primes.pop_back();
  reverse((primes).begin(), (primes).end());
  if (ans == 1) {
    if (!primes.empty()) {
      const int K = sqrt(primes.size());
      int left_numbers = primes.size() + 1;
      for (int block = 0; block * K < (int)primes.size(); ++block) {
        int R = min((int)primes.size(), (block + 1) * K);
        int L = block * K;
        for (int i = L; i < R; ++i) query('B', primes[i]);
        left_numbers -= R - L;
        if (query('A', 1) != left_numbers) {
          for (int i = L; i < R; ++i) {
            if (query('A', primes[i])) {
              ans = primes[i];
              break;
            }
          }
          break;
        }
      }
    }
  } else {
    for (int x : primes) {
      if (ans * (ll)x <= n && query('A', ans * x)) {
        ans *= x;
        break;
      }
    }
  }
  query('C', ans);
}
