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;
bool isPrime[100005], v[100005];
long long Prime[100005], cnt = 0;
void GetPrime(long long n) {
  memset(isPrime, 1, sizeof(isPrime));
  isPrime[1] = 0;
  for (long long i = 2; i <= n; i++) {
    if (isPrime[i]) Prime[++cnt] = i;
    for (long long j = 1; j <= cnt && i * Prime[j] <= n; j++) {
      isPrime[i * Prime[j]] = 0;
      if (i % Prime[j] == 0) break;
    }
  }
}
signed main() {
  long long n;
  cin >> n;
  GetPrime(n);
  long long k = sqrt(cnt), sum = n, ans = 1;
  bool flag = 0;
  for (long long i = 1; i <= cnt; i++) {
    if (i >= k && ans * Prime[i - k + 1] > n) break;
    cout << "B " << Prime[i] << endl;
    long long num = 0;
    for (long long j = Prime[i]; j <= n; j += Prime[i]) {
      if (!v[j]) {
        num++;
        sum--;
        v[j] = 1;
      }
    }
    long long x;
    cin >> x;
    if (x != num) {
      for (long long j = Prime[i]; j <= n; j *= Prime[i]) {
        cout << "A " << j << endl;
        cin >> x;
        if (x)
          ans *= Prime[i];
        else
          break;
      }
    }
    if ((i == cnt || i % k == 0) & !flag) {
      cout << "A 1" << endl;
      cin >> x;
      if (x != sum) {
        for (long long j = i - k + 1; j <= i; j++) {
          for (long long kk = Prime[j]; kk <= n; kk *= Prime[j]) {
            cout << "A " << kk << endl;
            cin >> x;
            if (x)
              ans *= Prime[j], flag = 1;
            else
              break;
          }
          if (flag) break;
        }
      }
    }
  }
  cout << "C " << ans << endl;
  return 0;
}