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;
const int MAXN = 100005;
int n;
bool is_prime[MAXN];
vector<int> primes;
void sieve(int n) {
  fill(is_prime + 2, is_prime + n + 1, true);
  for (int i = 2; i <= n; i++) {
    if (is_prime[i]) primes.push_back(i);
    for (auto p : primes) {
      if (p * i > n) break;
      is_prime[p * i] = false;
      if (i % p == 0) break;
    }
  }
}
bool vis[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  if (n == 1) {
    cout << "C 1" << endl;
    cout.flush();
    return 0;
  }
  sieve(n);
  int left = n, ans = 1, sqrt_n = sqrt(n);
  bool min_prim_flag = false;
  for (int i = 0; i < primes.size(); i++) {
    int p = primes[i];
    cout << "B " << p << endl;
    cout.flush();
    int mul_cnt = 0, real_mul_cnt;
    cin >> real_mul_cnt;
    for (int i = p; i <= n; i += p) {
      if (!vis[i]) {
        vis[i] = true;
        left--;
        mul_cnt++;
      }
    }
    if (mul_cnt != real_mul_cnt) {
      int res;
      for (long long j = p; j <= n; j *= p) {
        cout << "A " << j << endl;
        cout.flush();
        cin >> res;
        if (res != 1) break;
        ans *= p;
      }
    }
    if (((i + 1) % sqrt_n == 0 || i == primes.size() - 1) && !min_prim_flag) {
      cout << "A 1" << endl;
      cout.flush();
      int res;
      cin >> res;
      if (res != left) {
        for (int j = i - sqrt_n + 1; j <= i; j++) {
          cout << "A " << primes[j] << endl;
          cout.flush();
          cin >> res;
          if (res == 1) {
            for (long long k = primes[j]; k <= n; k *= primes[j]) {
              cout << "A " << k << endl;
              cout.flush();
              cin >> res;
              if (res != 1) break;
              ans *= primes[j];
            }
            min_prim_flag = true;
            break;
          }
        }
      }
    }
  }
  cout << "C " << ans << endl;
  cout.flush();
  return 0;
}