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;
const long long int MOD = 1e9 + 7;
vector<bool> isPrime;
void sieve(long long int n) {
  for (long long int i = 2; i <= n; i++) {
    if (isPrime[i]) {
      for (long long int j = 2 * i; j <= n; j += i) {
        isPrime[j] = false;
      }
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  isPrime.resize(n + 1, true);
  sieve(n);
  set<long long int> s;
  for (long long int i = 1; i <= n; i++) {
    s.insert(i);
  }
  long long int got;
  for (long long int i = 2; i * i <= n; i++) {
    if (!isPrime[i]) continue;
    cout << "B " << i << endl;
    for (long long int j = i; j <= n; j += i) {
      s.erase(j);
    }
    cin >> got;
  }
  cout << "A " << 1 << endl;
  cin >> got;
  if ((int)s.size() == got) {
    long long int ans = 1;
    vector<long long int> bag[100];
    long long int srt = sqrt(n);
    long long int c = 0;
    for (long long int i = srt + 1; i <= n; i++) {
      if (!isPrime[i]) continue;
      bag[c / 100].push_back(i);
      c++;
    }
    for (long long int i = 0; i <= c / 100; i++) {
      for (auto it : bag[i]) {
        cout << "B " << it << endl;
        cin >> got;
        s.erase(it);
      }
      cout << "A " << 1 << endl;
      cin >> got;
      if (got != (int)s.size()) {
        for (auto it : bag[i]) {
          cout << "A " << it << endl;
          cin >> got;
          if (got) ans *= it;
        }
        break;
      }
    }
    cout << "C " << ans << endl;
  } else {
    long long int ans = 1;
    for (long long int i = 2; i * i <= n; i++) {
      if (!isPrime[i]) continue;
      long long int val = i;
      while (val <= n) {
        cout << "A " << val << endl;
        cin >> got;
        if (got == 0) {
          val /= i;
          ans *= val;
          break;
        } else {
          val *= i;
          if (val > n) ans *= (val / i);
        }
      }
    }
    for (long long int i = sqrt(n) + 1; i <= n; i++) {
      if (!isPrime[i]) continue;
      cout << "B " << i << endl;
      cin >> got;
      if (got > 1) {
        ans *= i;
      }
    }
    cout << "C " << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
