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;
int query(char type, int a) {
  int res;
  cout << type << ' ' << a << endl;
  cin >> res;
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> cnt(n + 1, -1);
  for (int i = 2; i <= n; i++) {
    if (cnt[i] == -1) {
      cnt[i] = 1;
      for (int64_t j = 1LL * i * i; j <= n; j += i) {
        if (cnt[j] == -1) {
          cnt[j] = -2;
          cnt[i]++;
        }
      }
    }
  }
  vector<pair<int, int>> a;
  for (int i = 2; i <= n; i++) {
    if (cnt[i] > 0) {
      a.emplace_back(i, cnt[i]);
    }
  }
  bool found_first = false;
  int n_primes = a.size();
  int should_have = n;
  int batch = max(3, (int)sqrt(a.size()));
  int rep = (n_primes + batch - 1) / batch;
  int answer = 1;
  auto test_pow = [&](int target) -> int {
    if (1LL * target * target > n) return target;
    int x = target;
    while (x * target <= n) x *= target;
    while (query('A', x) == 0) {
      x /= target;
    }
    return x;
  };
  for (int i = 0; i < rep; i++) {
    int start = i * batch;
    int stop = min(n_primes, (i + 1) * batch);
    for (int j = start; j < stop; j++) {
      int res = query('B', a[j].first);
      if (res != a[j].second) {
        answer *= test_pow(a[j].first);
      }
      should_have -= a[j].second;
    }
    if (!found_first) {
      int now_remaining = query('A', 1);
      if (now_remaining != should_have) {
        for (int j = start; j < stop; j++) {
          int res = query('A', a[j].first);
          if (res != 0) {
            found_first = true;
            answer *= test_pow(a[j].first);
            break;
          }
        }
      }
    }
  }
  cout << "C " << answer << endl;
  return 0;
}